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4. Systems of Equations

4.4. Solve Applications with Systems of Equations

Lynn Marecek and MaryAnne Anthony-Smith

Learning Objectives

By the end of this section it is expected that you will be able to:

  • Translate to a system of equations
  • Solve direct translation applications
  • Solve geometry applications
  • Solve uniform motion applications

Previously in this chapter we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem Solving Strategy for Systems of Linear Equations.

Use a problem solving strategy for systems of linear equations.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

Translate to a System of Equations

Many of the problems we solved in earlier applications related two quantities.

Let’s see how we can translate these problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem Solving Strategy.

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

This figure has four rows and three columns. The first row reads, “Step 1: Read the problem. Make sure you understand all the words and ideas. This is a number problem. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.”

The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

\left\{\begin{array}{c}m+n=-23\hfill \\ m=n-7\hfill \end{array}

We’ll do another example where we stop after we write the system of equations.

A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

A couple has a total household income of $84,000. The husband earns $18,000 less than twice what the wife earns. How much does the wife earn?

\left\{\begin{array}{c}w+h=84,000\hfill \\ h=2w-18,000\hfill \end{array}

Solve Direct Translation Applications

We set up, but did not solve, the systems of equations in examples 1 and 2. Now we’ll translate a situation to a system of equations and then solve it.

Translate to a system of equations and then solve:

Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

Ali is 12 years older than his youngest sister, Jameela. The sum of their ages is 40. Find their ages.

Ali is 28 and Jameela is 16.

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn for each minute of circuit training?

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Solve Geometry Applications

We solved geometry applications using properties of triangles and rectangles. Now we’ll add to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

Complementary and Supplementary Angles

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

The angle measures are 55 degrees and 35 degrees.

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

The angle measures are 42 degrees and 138 degrees.

Randall has 125 feet of fencing to enclose the rectangular part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

The length is 60 feet and the width is 35 feet.

Solve Uniform Motion Applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was D = rt where D is the distance travelled, r is the rate, and t is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”

Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

It will take Clark 4 hours to catch Mitchell.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c .

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is b + c .

This figure shows a boat floating in water. On the right, there is an arrow pointing towards the boat. It is labeled “c.” On the left, there is an arrow pointing away from the boat. It is labeled “b.”

We’ll put some numbers to this situation in the next example.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Read  the problem.

This is a uniform motion problem and a picture will help us visualize the situation.

This figure shows an arrow labeled “c” which continues to the right, representing the wave. Under the wave is a ray that points to the right and is labeled “four hours.” Under this ray is another ray pointing to the left labeled “five hours.” It is the same length as the ray labeled “four hours.” There is a bracket under the ray labeled “five hours.” The bracket is labeled “60 miles.”

Translate to a system of equations and then solve: Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

A private jet can fly 1095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

This is a uniform motion problem and a picture will help us visualize.

This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”

Translate to a system of equations and then solve: A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1025 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

4.4 Exercise Set

In the following exercises, translate to a system of equations and solve the system.

  • The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.
  • The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.
  • Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.
  • Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.
  • A married couple together earn ?75,000. The husband earns ?15,000 more than five times what his wife earns. What does the wife earn?
  • Daniela invested a total of ?50,000, some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was ?5000 more than twice the amount she put into the CD. How much did she invest in each account?
  • In her last two years in college, Marlene received? 42,000 in loans. The first year she received a loan that was ?6,000 less than three times the amount of the second year’s loan. What was the amount of her loan for each year?
  • Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.
  • The age of Noelle’s dad is six less than three times Noelle’s age. The sum of their ages is seventy-four. Find their ages.
  • Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?
  • Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?
  • Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for $116. Lisa bought two notebooks and three thumb dives for $68. Find the cost of each notebook and each thumb drive.

In the following exercises, translate to a system of equations and solve.

  • The difference of two complementary angles is 30 degrees. Find the measures of the angles.
  • The difference of two supplementary angles is 70 degrees. Find the measures of the angles.
  • The difference of two supplementary angles is 8 degrees. Find the measures of the angles.
  • The difference of two complementary angles is 55 degrees. Find the measures of the angles.
  • Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.
  • Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.
  • Wayne is hanging a string of lights 45 feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.
  • A frame around a rectangular family portrait has a perimeter of 60 inches. The length is fifteen less than twice the width. Find the length and width of the frame.
  • Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?
  • At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?
  • The Jones family took a 12 mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.
  • A motor boat traveled 18 miles down a river in two hours but going back upstream, it took 4.5 hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.).
  • A small jet can fly 1,072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
  • A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.  

\left\{\begin{array}{c}s+a=425\hfill \\ 5s+8a=2,851\hfill \end{array}

  • The numbers are 6 and 9.
  • The numbers are −5 and −25.
  • The numbers are 5 and 4.
  • The numbers are 2 and 3.
  • $10,000
  • She put $15,000 into a CD and $35,000 in bonds.
  • The amount of the first year’s loan was $30,000 and the amount of the second year’s loan was $12,000.
  • Bethany is 16 years old and Alyssa is 28 years old.
  • Noelle is 20 years old and her dad is 54 years old.
  • The small container holds 20 gallons and the large container holds 30 gallons.
  • There were 10 calories burned jogging and 10 calories burned cycling.
  • Notebooks are $4 and thumb drives are $20.
  • The measures are 60 degrees and 30 degrees.
  • The measures are 125 degrees and 55 degrees.
  • 94 degrees and 86 degrees
  • 72.5 degrees and 17.5 degrees
  • The measures are 44 degrees and 136 degrees.
  • The measures are 34 degrees and 56 degrees.
  • The width is 10 feet and the length is 25 feet.
  • The width is 15 feet and the length is 15 feet.
  • It took Sarah’s sister 12 hours.
  • It took Lucy’s friend 2 hours.
  • The canoe rate is 5 mph and the current rate is 1 mph.
  • The boat rate is 6.5 mph and the current rate is 2.5 mph.
  • The jet rate is 240 mph and the wind speed is 28 mph.
  • The jet rate is 415 mph and the wind speed is 19 mph.

s=183,a=242

Business/Technical Mathematics by Lynn Marecek and MaryAnne Anthony-Smith is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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using systems of two equations to solve application problems

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Systems of Linear Equations

Solve Applications with Systems of Equations

Learning objectives.

By the end of this section, you will be able to:

  • Solve direct translation applications
  • Solve geometry applications

Solve uniform motion applications

Before you get started, take this readiness quiz.

Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

  • Read the problem. Make sure all the words and ideas are understood.
  • Identify what we are looking for.
  • Name what we are looking for. Choose variables to represent those quantities.
  • Translate into a system of equations.
  • Solve the system of equations using good algebra techniques.
  • Check the answer in the problem and make sure it makes sense.
  • Answer the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

-6.

Geraldine has been offered positions by two insurance companies. The first company pays a salary of ?12,000 plus a commission of ?100 for each policy sold. The second pays a salary of ?20,000 plus a commission of ?50 for each policy sold. How many policies would need to be sold to make the total pay the same?

160 policies

Kenneth currently sells suits for company A at a salary of ?22,000 plus a ?10 commission for each suit sold. Company B offers him a position with a salary of ?28,000 plus a ?4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Translate to a system of equations and then solve:

When Jenna spent 10 minutes on the elliptical trainer and then did circuit training for 20 minutes, her fitness app says she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

Erin burned 11 calories for each minute on the rowing machine and 5 calories for each minute of weight lifting.

Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two complementary angles add to 90 degrees. The measures of two supplementary angles add to 180 degrees.

Two angles are complementary if the sum of the measures of their angles is 90 degrees.

Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

If two angles are complementary, we say that one angle is the complement of the other.

If two angles are supplementary, we say that one angle is the supplement of the other.

Translate to a system of equations and then solve.

The difference of two complementary angles is 26 degrees. Find the measures of the angles.

\begin{array}{cccc}\mathbf{\text{Step 1. Read}}\phantom{\rule{0.2em}{0ex}}\text{the problem.}\hfill & & & \\ \mathbf{\text{Step 2. Identify}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & \begin{array}{c}\text{We are looking for the measure of each}\hfill \\ \text{angle.}\hfill \end{array}\hfill \\ \mathbf{\text{Step 3. Name}}\phantom{\rule{0.2em}{0ex}}\text{what we are looking for.}\hfill & & & \begin{array}{c}\text{Let}\phantom{\rule{0.2em}{0ex}}x=\phantom{\rule{0.2em}{0ex}}\text{the measure of the first angle.}\hfill \\   y=\phantom{\rule{0.2em}{0ex}}\text{the measure of the second angle}\hfill \end{array}\hfill \\ \begin{array}{c}\mathbf{\text{Step 4. Translate}}\phantom{\rule{0.2em}{0ex}}\text{into a system of}\hfill \\ \text{equations.}\hfill \end{array}\hfill & & & \begin{array}{c}\text{The angles are complementary.}\hfill \\ \phantom{\rule{5.5em}{0ex}}x+y=90\hfill \end{array}\hfill \\ & & & \begin{array}{c}\text{The difference of the two angles is 26}\hfill \\ \text{degrees.}\hfill \end{array}\hfill \\ & & & \phantom{\rule{5.6em}{0ex}}x-y=26\hfill \\ \\ \\ \text{The system is shown.}\hfill & & & \phantom{\rule{5.3em}{0ex}}\left\{\begin{array}{c}x+y=90\hfill \\ x-y=26\hfill \end{array}\hfill \\ \\ \\ \begin{array}{c}\mathbf{\text{Step 5. Solve}}\phantom{\rule{0.2em}{0ex}}\text{the system of equations}\hfill \\ \text{by elimination.}\hfill \end{array}\hfill & & & \phantom{\rule{5em}{0ex}}\begin{array}{c}\underset{___________}{\left\{\begin{array}{c}x+y=90\hfill \\ x-y=26\hfill \end{array}}\hfill \\  2x\phantom{\rule{1em}{0ex}}=116\hfill \end{array}\hfill \\ \text{Substitute}\phantom{\rule{0.2em}{0ex}}x=58\phantom{\rule{0.2em}{0ex}}\text{into the first equation.}\hfill & & & \phantom{\rule{5.2em}{0ex}}\begin{array}{c}\hfill x=58\\ \\ \\ \hfill x+y=90\\ \hfill 58+y=90\\ \hfill y=32\end{array}\hfill \\ \begin{array}{c}\mathbf{\text{Step 6. Check}}\phantom{\rule{0.2em}{0ex}}\text{the answer in the problem.}\hfill \\ \\ \\ \\ \phantom{\rule{4em}{0ex}}\begin{array}{c}58+32=90✓\hfill \\ 58-32=26✓\hfill \end{array}\hfill \end{array}\hfill & & & \\ \mathbf{\text{Step 7. Answer}}\phantom{\rule{0.2em}{0ex}}\text{the question.}\hfill & & & \text{The angle measures are 58 and 32 degrees.}\hfill \end{array}

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

The angle measures are 55 and 35.

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

The angle measures are 5 and 85.

In the next example, we remember that the measures of supplementary angles add to 180.

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

The angle measures are 42 and 138.

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

The angle measures are 66 and 114.

Recall that the angles of a triangle add up to 180 degrees. A right triangle has one angle that is 90 degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Randall has 125 feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be 5 feet more than four times as long as the width. Find the length and the width.

Mario wants to put a fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

The length is 60 feet and the width is 35 feet.

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

The length is 60 feet and the width is 38 feet.

D=rt

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Joni left St. Louis on the interstate, driving west towards Denver at a speed of 65 miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving 78 miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

.

Identify and name what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, k , and Joni, j , will each drive.

.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

\phantom{\rule{5em}{0ex}}65j=78k

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

It will take Clark 4 hours to catch Mitchell.

\left(\frac{1}{4}\phantom{\rule{0.2em}{0ex}}\text{hour}\right)

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water b and the speed of the river current c .

b+c.

We’ll put some numbers to this situation in the next example.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a tailwind . A wind current blowing against the direction of the plane is called a headwind .

A private jet can fly 1,095 miles in three hours with a tailwind but only 987 miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

Access this online resource for additional instruction and practice with systems of equations.

  • Systems of Equations

Key Concepts

Practice makes perfect.

Direct Translation Applications

In the following exercises, translate to a system of equations and solve.

The sum of two number is 15. One number is 3 less than the other. Find the numbers.

The sum of two number is 30. One number is 4 less than the other. Find the numbers.

The sum of two number is −16. One number is 20 less than the other. Find the numbers.

-26.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

-27.

Maxim has been offered positions by two car companies. The first company pays a salary of ?10,000 plus a commission of ?1000 for each car sold. The second pays a salary of ?20,000 plus a commission of ?500 for each car sold. How many cars would need to be sold to make the total pay the same?

Jackie has been offered positions by two cable companies. The first company pays a salary of ?14,000 plus a commission of ?100 for each cable package sold. The second pays a salary of ?20,000 plus a commission of ?25 for each cable package sold. How many cable packages would need to be sold to make the total pay the same?

Eighty cable packages would need to be sold to make the total pay the same.

Amara currently sells televisions for company A at a salary of ?17,000 plus a ?100 commission for each television she sells. Company B offers her a position with a salary of ?29,000 plus a ?20 commission for each television she sells. How televisions would Amara need to sell for the options to be equal?

Mitchell currently sells stoves for company A at a salary of ?12,000 plus a ?150 commission for each stove he sells. Company B offers him a position with a salary of ?24,000 plus a ?50 commission for each stove he sells. How many stoves would Mitchell need to sell for the options to be equal?

Mitchell would need to sell 120 stoves for the companies to be equal.

Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?

June needs 48 gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

8 and 40 gallons

Shelly spent 10 minutes jogging and 20 minutes cycling and burned 300 calories. The next day, Shelly swapped times, doing 20 minutes of jogging and 10 minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?

Drew burned 1800 calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned 3200 calories. How many calories did he burn per hour when playing basketball? How many calories did he burn per hour when canoeing?

1000 calories playing basketball and 400 calories canoeing

Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for ?116. Lisa bought two notebooks and three thumb dives for ?68. Find the cost of each notebook and each thumb drive.

Nancy bought seven pounds of oranges and three pounds of bananas for ?17. Her husband later bought three pounds of oranges and six pounds of bananas for ?12. What was the cost per pound of the oranges and the bananas?

Oranges cost ?2 per pound and bananas cost ?1 per pound

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for ?114 or she is able to buy 2 shirts and 4 sweaters for ?164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for ?40 or he is able to buy 5 packages of paper and 6 staplers for ?62. How much does a package of paper cost? How much does a stapler cost?

Package of paper ?4, stapler ?7

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

Hot dog 150 calories, cup of cottage cheese 220 calories

Molly is making strawberry infused water. For each ounce of strawberry juice, she uses three times as many ounces of water as juice. How many ounces of strawberry juice and how many ounces of water does she need to make 64 ounces of strawberry infused water?

Owen is making lemonade from concentrate. The number of quarts of water he needs is 4 times the number of quarts of concentrate. How many quarts of water and how many quarts of concentrate does Owen need to make 100 quarts of lemonade?

Owen will need 80 quarts of water and 20 quarts of concentrate to make 100 quarts of lemonade.

The difference of two complementary angles is 55 degrees. Find the measures of the angles.

The difference of two complementary angles is 17 degrees. Find the measures of the angles.

53.5

Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.

Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

16 degrees and 74 degrees

The difference of two supplementary angles is 8 degrees. Find the measures of the angles.

The difference of two supplementary angles is 88 degrees. Find the measures of the angles.

134 degrees and 46 degrees

Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.

Two angles are supplementary. The measure of the larger angle is five less than four times the measure of the smaller angle. Find the measures of both angles.

37 degrees and 143 degrees

The measure of one of the small angles of a right triangle is 14 more than 3 times the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 26 more than 3 times the measure of the other small angle. Find the measure of both angles.

16\text{°}

The measure of one of the small angles of a right triangle is 15 less than twice the measure of the other small angle. Find the measure of both angles.

The measure of one of the small angles of a right triangle is 45 less than twice the measure of the other small angle. Find the measure of both angles.

45\text{°}

Wayne is hanging a string of lights 45 feet long around the three sides of his patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.

Darrin is hanging 200 feet of Christmas garland on the three sides of fencing that enclose his front yard. The length is five feet less than three times the width. Find the length and width of the fencing.

Width is 41 feet and length is 118 feet.

A frame around a family portrait has a perimeter of 90 inches. The length is fifteen less than twice the width. Find the length and width of the frame.

The perimeter of a toddler play area is 100 feet. The length is ten more than three times the width. Find the length and width of the play area.

Width is 10 feet and length is 40 feet.

Solve Uniform Motion Applications

Sarah left Minneapolis heading east on the interstate at a speed of 60 mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of 70 mph. How long will it take for Sarah’s sister to catch up to Sarah?

College roommates John and David were driving home to the same town for the holidays. John drove 55 mph, and David, who left an hour later, drove 60 mph. How long will it take for David to catch up to John?

At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of 40 mph. Lucy’s friend left the beach for home 30 minutes (half an hour) later, and drove 50 mph. How long did it take Lucy’s friend to catch up to Lucy?

Felecia left her home to visit her daughter driving 45 mph. Her husband waited for the dog sitter to arrive and left home twenty minutes (1/3 hour) later. He drove 55 mph to catch up to Felecia. How long before he reaches her?

1.5

The Jones family took a 12-mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.

A motor boat travels 60 miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

Boat rate is 16 mph and current rate is 4 mph.

4.5

A river cruise boat sailed 80 miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current.

Boat rate is 18 mph and current rate is 2 mph.

A small jet can fly 1072 miles in 4 hours with a tailwind but only 848 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A small jet can fly 1435 miles in 5 hours with a tailwind but only 1,215 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Jet rate is 265 mph and wind speed is 22 mph.

A commercial jet can fly 868 miles in 2 hours with a tailwind but only 792 miles in 2 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

A commercial jet can fly 1,320 miles in 3 hours with a tailwind but only 1170 miles in 3 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Jet rate is 415 mph and wind speed is 25 mph.

Writing Exercises

Write an application problem similar to (Figure) . Then translate to a system of equations and solve it.

Write a uniform motion problem similar to (Figure) that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.

Answers will vary.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

This table has 4 columns, 3 rows and a header row. The header row labels each column: I can, confidently, with some help and no, I don’t get it. The first column has the following statements: solve direct translation applications, solve geometry applications, solve uniform motion applications. The remaining columns are empty.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

Intermediate Algebra by OSCRiceUniversity is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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Solve Applications with Systems of Equations

4.2 solve applications with systems of equations.

Topics covered in this section are:

  • Solve direct translation applications
  • Solve geometry applications
  • Solve uniform motion applications

4.2.1 Solve Direct Translation Applications

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

HOW TO: Solve applications with systems of equations.

  • Read  the problem. Make sure all the words and ideas are understood.
  • Identify  what we are looking for.
  • Name  what we are looking for. Choose variables to represent those quantities.
  • Translate  into a system of equations.
  • Solve  the system of equations using good algebra techniques.
  • Check  the answer in the problem and make sure it makes sense.
  • Answer  the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $\$25,000$ plus $\$15$ for each training session. Option B would pay her $\$10,000+\$40$  for each training session. How many training sessions would make the salary options equal?

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

Translate to a system of equations and then solve: When Jenna spent $10$ minutes on the elliptical trainer and then did circuit training for $20$ minutes, her fitness app says she burned $278$ calories. When she spent $20$ minutes on the elliptical trainer and $30$ minutes circuit training she burned $473$ calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

4.2.2 Solve Geometry Applications

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two  complementary angles  add to $90$ degrees. The measures of two  supplementary angles  add to $180$ degrees.

COMPLEMENTARY AND SUPPLEMENTARY ANGLES

Two angles are complementary if the sum of the measures of their angles is $90$ degrees.

Two angles are supplementary if the sum of the measures of their angles is $180$ degrees.

If two angles are complementary, we say that  one angle is the complement of the other.

If two angles are supplementary, we say that  one angle is the supplement of the other.

Translate to a system of equations and then solve. The difference of two complementary angles is $26$ degrees. Find the measures of the angles.

In the next example, we remember that the measures of supplementary angles add to $180$.

Translate to a system of equations and then solve: Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

Recall that the angles of a triangle add up to $180$ degrees. A right triangle has one angle that is $90$ degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

We will draw and label a figure.

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

Translate to a system of equations and then solve: Randall has $125$ feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be $5$ feet more than four times as long as the width. Find the length and the width.

4.2.3 Solve uniform motion applications

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was $D=rt$ where $D$ is the distance traveled, $r$ is the rate, and $t$ is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

Translate to a system of equations and then solve: Joni left St. Louis on the interstate, driving west towards Denver at a speed of $65$ miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving $78$ miles per hour. How long will it take Kelly to catch up to Joni?

A diagram is useful in helping us visualize the situation.

This figure helps visualize the situation. Joni is driving west from St, Louis toward Denver at 65 miles per hour. Kelly leaves half an hour after Joni on the same route but driving 78 miles per hour.

Identify and name  what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, $k$, and Joni, $j$, will each drive.

Since $D=r \cdot t$ we can fill in the Distance column.

Translate into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

Also, since Kelly left later, her time will be $\frac{1}{2}$ hour less than Joni’s time. So,

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water $b$ and the speed of the river current $c$.

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving $b+c$.

The figure shows a boat and two vertical arrows, both pointing up. The one to the left of the boat is the river current, c,  and the one to the right is the speed of the boat, b.

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is $b−c$.

The figure shows a boat and two vertical arrows. The one to the right of the boat is labeled b, and points up. The one to the left of the boat is labeled c and points down. This figure shows the boat going against the current.

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve. A river cruise ship sailed $60$ miles downstream for $4$ hours and then took $5$ hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a  tailwind . A wind current blowing against the direction of the plane is called a  headwind .

Translate to a system of equations and then solve: A private jet can fly $1,095$ miles in three hours with a tailwind but only $987$ miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

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  • Revision and Adaption. Provided by: Minute Math. License:  CC BY 4.0

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  • Marecek, L., & Mathis, A. H. (2020). Solve Applications with Systems of Equations. In Intermediate Algebra 2e. OpenStax.  https://openstax.org/books/intermediate-algebra-2e/pages/4-2-solve-applications-with-systems-of-equations .  License:  CC BY 4.0 . Access for free at  https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction

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Unit 6: Systems of equations

About this unit, introduction to systems of equations.

  • Systems of equations: trolls, tolls (1 of 2) (Opens a modal)
  • Systems of equations: trolls, tolls (2 of 2) (Opens a modal)
  • Testing a solution to a system of equations (Opens a modal)
  • Systems of equations with graphing: y=7/5x-5 & y=3/5x-1 (Opens a modal)
  • Systems of equations with graphing: exact & approximate solutions (Opens a modal)
  • Setting up a system of equations from context example (pet weights) (Opens a modal)
  • Setting up a system of linear equations example (weight and price) (Opens a modal)
  • Interpreting points in context of graphs of systems (Opens a modal)
  • Solutions of systems of equations Get 3 of 4 questions to level up!
  • Systems of equations with graphing Get 3 of 4 questions to level up!
  • Creating systems in context Get 3 of 4 questions to level up!
  • Interpret points relative to a system Get 3 of 4 questions to level up!

Solving systems of equations with substitution

  • Systems of equations with substitution: potato chips (Opens a modal)
  • Systems of equations with substitution: -3x-4y=-2 & y=2x-5 (Opens a modal)
  • Substitution method review (systems of equations) (Opens a modal)
  • Systems of equations with substitution Get 3 of 4 questions to level up!

Solving systems of equations with elimination

  • Systems of equations with elimination: King's cupcakes (Opens a modal)
  • Elimination strategies (Opens a modal)
  • Systems of equations with elimination: x-4y=-18 & -x+3y=11 (Opens a modal)
  • Systems of equations with elimination: potato chips (Opens a modal)
  • Systems of equations with elimination (and manipulation) (Opens a modal)
  • Elimination method review (systems of linear equations) (Opens a modal)
  • Combining equations Get 3 of 4 questions to level up!
  • Elimination strategies Get 3 of 4 questions to level up!
  • Systems of equations with elimination Get 3 of 4 questions to level up!
  • Systems of equations with elimination challenge Get 3 of 4 questions to level up!

Equivalent systems of equations

  • Why can we subtract one equation from the other in a system of equations? (Opens a modal)
  • Worked example: equivalent systems of equations (Opens a modal)
  • Worked example: non-equivalent systems of equations (Opens a modal)
  • Reasoning with systems of equations (Opens a modal)
  • Equivalent systems of equations review (Opens a modal)
  • Reasoning with systems of equations Get 3 of 4 questions to level up!

Number of solutions to systems of equations

  • Systems of equations number of solutions: fruit prices (1 of 2) (Opens a modal)
  • Systems of equations number of solutions: fruit prices (2 of 2) (Opens a modal)
  • Solutions to systems of equations: consistent vs. inconsistent (Opens a modal)
  • Solutions to systems of equations: dependent vs. independent (Opens a modal)
  • Number of solutions to a system of equations (Opens a modal)
  • Number of solutions to a system of equations graphically (Opens a modal)
  • Number of solutions to a system of equations algebraically (Opens a modal)
  • How many solutions does a system of linear equations have if there are at least two? (Opens a modal)
  • Number of solutions to system of equations review (Opens a modal)
  • Number of solutions to a system of equations graphically Get 3 of 4 questions to level up!
  • Number of solutions to a system of equations algebraically Get 3 of 4 questions to level up!

Systems of equations word problems

  • Age word problem: Imran (Opens a modal)
  • Age word problem: Ben & William (Opens a modal)
  • Age word problem: Arman & Diya (Opens a modal)
  • System of equations word problem: walk & ride (Opens a modal)
  • System of equations word problem: no solution (Opens a modal)
  • System of equations word problem: infinite solutions (Opens a modal)
  • Systems of equations with elimination: TV & DVD (Opens a modal)
  • Systems of equations with elimination: apples and oranges (Opens a modal)
  • Systems of equations with substitution: coins (Opens a modal)
  • Systems of equations with elimination: coffee and croissants (Opens a modal)
  • Systems of equations: FAQ (Opens a modal)
  • Age word problems Get 3 of 4 questions to level up!
  • Systems of equations word problems Get 3 of 4 questions to level up!
  • Systems of equations word problems (with zero and infinite solutions) Get 3 of 4 questions to level up!
  • 4.1 Solve Systems of Linear Equations with Two Variables
  • Introduction
  • 1.1 Use the Language of Algebra
  • 1.2 Integers
  • 1.3 Fractions
  • 1.4 Decimals
  • 1.5 Properties of Real Numbers
  • Key Concepts
  • Review Exercises
  • Practice Test
  • 2.1 Use a General Strategy to Solve Linear Equations
  • 2.2 Use a Problem Solving Strategy
  • 2.3 Solve a Formula for a Specific Variable
  • 2.4 Solve Mixture and Uniform Motion Applications
  • 2.5 Solve Linear Inequalities
  • 2.6 Solve Compound Inequalities
  • 2.7 Solve Absolute Value Inequalities
  • 3.1 Graph Linear Equations in Two Variables
  • 3.2 Slope of a Line
  • 3.3 Find the Equation of a Line
  • 3.4 Graph Linear Inequalities in Two Variables
  • 3.5 Relations and Functions
  • 3.6 Graphs of Functions
  • 4.2 Solve Applications with Systems of Equations
  • 4.3 Solve Mixture Applications with Systems of Equations
  • 4.4 Solve Systems of Equations with Three Variables
  • 4.5 Solve Systems of Equations Using Matrices
  • 4.6 Solve Systems of Equations Using Determinants
  • 4.7 Graphing Systems of Linear Inequalities
  • 5.1 Add and Subtract Polynomials
  • 5.2 Properties of Exponents and Scientific Notation
  • 5.3 Multiply Polynomials
  • 5.4 Dividing Polynomials
  • Introduction to Factoring
  • 6.1 Greatest Common Factor and Factor by Grouping
  • 6.2 Factor Trinomials
  • 6.3 Factor Special Products
  • 6.4 General Strategy for Factoring Polynomials
  • 6.5 Polynomial Equations
  • 7.1 Multiply and Divide Rational Expressions
  • 7.2 Add and Subtract Rational Expressions
  • 7.3 Simplify Complex Rational Expressions
  • 7.4 Solve Rational Equations
  • 7.5 Solve Applications with Rational Equations
  • 7.6 Solve Rational Inequalities
  • 8.1 Simplify Expressions with Roots
  • 8.2 Simplify Radical Expressions
  • 8.3 Simplify Rational Exponents
  • 8.4 Add, Subtract, and Multiply Radical Expressions
  • 8.5 Divide Radical Expressions
  • 8.6 Solve Radical Equations
  • 8.7 Use Radicals in Functions
  • 8.8 Use the Complex Number System
  • 9.1 Solve Quadratic Equations Using the Square Root Property
  • 9.2 Solve Quadratic Equations by Completing the Square
  • 9.3 Solve Quadratic Equations Using the Quadratic Formula
  • 9.4 Solve Equations in Quadratic Form
  • 9.5 Solve Applications of Quadratic Equations
  • 9.6 Graph Quadratic Functions Using Properties
  • 9.7 Graph Quadratic Functions Using Transformations
  • 9.8 Solve Quadratic Inequalities
  • 10.1 Finding Composite and Inverse Functions
  • 10.2 Evaluate and Graph Exponential Functions
  • 10.3 Evaluate and Graph Logarithmic Functions
  • 10.4 Use the Properties of Logarithms
  • 10.5 Solve Exponential and Logarithmic Equations
  • 11.1 Distance and Midpoint Formulas; Circles
  • 11.2 Parabolas
  • 11.3 Ellipses
  • 11.4 Hyperbolas
  • 11.5 Solve Systems of Nonlinear Equations
  • 12.1 Sequences
  • 12.2 Arithmetic Sequences
  • 12.3 Geometric Sequences and Series
  • 12.4 Binomial Theorem

Learning Objectives

By the end of this section, you will be able to:

  • Determine whether an ordered pair is a solution of a system of equations
  • Solve a system of linear equations by graphing
  • Solve a system of equations by substitution
  • Solve a system of equations by elimination
  • Choose the most convenient method to solve a system of linear equations

Be Prepared 4.1

Before you get started, take this readiness quiz.

For the equation y = 2 3 x − 4 , y = 2 3 x − 4 , ⓐ Is ( 6 , 0 ) ( 6 , 0 ) a solution? ⓑ Is ( −3 , −2 ) ( −3 , −2 ) a solution? If you missed this problem, review Example 3.2 .

Be Prepared 4.2

Find the slope and y -intercept of the line 3 x − y = 12 . 3 x − y = 12 . If you missed this problem, review Example 3.16 .

Be Prepared 4.3

Find the x- and y -intercepts of the line 2 x − 3 y = 12 . 2 x − 3 y = 12 . If you missed this problem, review Example 3.8 .

Determine Whether an Ordered Pair is a Solution of a System of Equations

In Solving Linear Equations , we learned how to solve linear equations with one variable. Now we will work with two or more linear equations grouped together, which is known as a system of linear equations .

System of Linear Equations

When two or more linear equations are grouped together, they form a system of linear equations .

In this section, we will focus our work on systems of two linear equations in two unknowns. We will solve larger systems of equations later in this chapter.

An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations.

A linear equation in two variables, such as 2 x + y = 7 , 2 x + y = 7 , has an infinite number of solutions. Its graph is a line. Remember, every point on the line is a solution to the equation and every solution to the equation is a point on the line.

To solve a system of two linear equations, we want to find the values of the variables that are solutions to both equations. In other words, we are looking for the ordered pairs ( x , y ) ( x , y ) that make both equations true. These are called the solutions of a system of equations .

Solutions of a System of Equations

The solutions of a system of equations are the values of the variables that make all the equations true. A solution of a system of two linear equations is represented by an ordered pair ( x , y ) . ( x , y ) .

To determine if an ordered pair is a solution to a system of two equations, we substitute the values of the variables into each equation. If the ordered pair makes both equations true, it is a solution to the system.

Example 4.1

Determine whether the ordered pair is a solution to the system { x − y = −1 2 x − y = −5 . { x − y = −1 2 x − y = −5 .

ⓐ ( −2 , −1 ) ( −2 , −1 ) ⓑ ( −4 , −3 ) ( −4 , −3 )

Determine whether the ordered pair is a solution to the system { 3 x + y = 0 x + 2 y = −5 . { 3 x + y = 0 x + 2 y = −5 .

ⓐ ( 1 , −3 ) ( 1 , −3 ) ⓑ ( 0 , 0 ) ( 0 , 0 )

Determine whether the ordered pair is a solution to the system { x − 3 y = −8 − 3 x − y = 4 . { x − 3 y = −8 − 3 x − y = 4 .

ⓐ ( 2 , −2 ) ( 2 , −2 ) ⓑ ( −2 , 2 ) ( −2 , 2 )

Solve a System of Linear Equations by Graphing

In this section, we will use three methods to solve a system of linear equations. The first method we’ll use is graphing.

The graph of a linear equation is a line. Each point on the line is a solution to the equation. For a system of two equations, we will graph two lines. Then we can see all the points that are solutions to each equation. And, by finding what the lines have in common, we’ll find the solution to the system.

Most linear equations in one variable have one solution, but we saw that some equations, called contradictions, have no solutions and for other equations, called identities, all numbers are solutions.

Similarly, when we solve a system of two linear equations represented by a graph of two lines in the same plane, there are three possible cases, as shown.

Each time we demonstrate a new method, we will use it on the same system of linear equations. At the end of the section you’ll decide which method was the most convenient way to solve this system.

Example 4.2

How to solve a system of equations by graphing.

Solve the system by graphing { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Solve the system by graphing: { x − 3 y = −3 x + y = 5 . { x − 3 y = −3 x + y = 5 .

Solve the system by graphing: { − x + y = 1 3 x + 2 y = 12 . { − x + y = 1 3 x + 2 y = 12 .

The steps to use to solve a system of linear equations by graphing are shown here.

Solve a system of linear equations by graphing.

  • Step 1. Graph the first equation.
  • Step 2. Graph the second equation on the same rectangular coordinate system.
  • Step 3. Determine whether the lines intersect, are parallel, or are the same line.
  • If the lines intersect, identify the point of intersection. This is the solution to the system.
  • If the lines are parallel, the system has no solution.
  • If the lines are the same, the system has an infinite number of solutions.
  • Step 5. Check the solution in both equations.

In the next example, we’ll first re-write the equations into slope–intercept form as this will make it easy for us to quickly graph the lines.

Example 4.3

Solve the system by graphing: { 3 x + y = − 1 2 x + y = 0 . { 3 x + y = − 1 2 x + y = 0 .

We’ll solve both of these equations for y y so that we can easily graph them using their slopes and y -intercepts.

Solve the system by graphing: { − x + y = 1 2 x + y = 10 . { − x + y = 1 2 x + y = 10 .

Solve the system by graphing: { 2 x + y = 6 x + y = 1 . { 2 x + y = 6 x + y = 1 .

In all the systems of linear equations so far, the lines intersected and the solution was one point. In the next two examples, we’ll look at a system of equations that has no solution and at a system of equations that has an infinite number of solutions.

Example 4.4

Solve the system by graphing: { y = 1 2 x − 3 x − 2 y = 4 . { y = 1 2 x − 3 x − 2 y = 4 .

Solve the system by graphing: { y = − 1 4 x + 2 x + 4 y = − 8 . { y = − 1 4 x + 2 x + 4 y = − 8 .

Solve the system by graphing: { y = 3 x − 1 6 x − 2 y = 6 . { y = 3 x − 1 6 x − 2 y = 6 .

Sometimes the equations in a system represent the same line. Since every point on the line makes both equations true, there are infinitely many ordered pairs that make both equations true. There are infinitely many solutions to the system.

Example 4.5

Solve the system by graphing: { y = 2 x − 3 − 6 x + 3 y = − 9 . { y = 2 x − 3 − 6 x + 3 y = − 9 .

If you write the second equation in slope-intercept form, you may recognize that the equations have the same slope and same y -intercept.

Solve the system by graphing: { y = − 3 x − 6 6 x + 2 y = − 12 . { y = − 3 x − 6 6 x + 2 y = − 12 .

Try It 4.10

Solve the system by graphing: { y = 1 2 x − 4 2 x − 4 y = 16 . { y = 1 2 x − 4 2 x − 4 y = 16 .

When we graphed the second line in the last example, we drew it right over the first line. We say the two lines are coincident . Coincident lines have the same slope and same y- intercept.

Coincident Lines

Coincident lines have the same slope and same y- intercept.

The systems of equations in Example 4.2 and Example 4.3 each had two intersecting lines. Each system had one solution.

In Example 4.5 , the equations gave coincident lines, and so the system had infinitely many solutions.

The systems in those three examples had at least one solution. A system of equations that has at least one solution is called a consistent system.

A system with parallel lines, like Example 4.4 , has no solution. We call a system of equations like this inconsistent. It has no solution.

Consistent and Inconsistent Systems

A consistent system of equations is a system of equations with at least one solution.

An inconsistent system of equations is a system of equations with no solution.

We also categorize the equations in a system of equations by calling the equations independent or dependent . If two equations are independent, they each have their own set of solutions. Intersecting lines and parallel lines are independent.

If two equations are dependent, all the solutions of one equation are also solutions of the other equation. When we graph two dependent equations, we get coincident lines.

Let’s sum this up by looking at the graphs of the three types of systems. See below and Table 4.1 .

Example 4.6

Without graphing, determine the number of solutions and then classify the system of equations.

ⓐ { y = 3 x − 1 6 x − 2 y = 12 { y = 3 x − 1 6 x − 2 y = 12 ⓑ { 2 x + y = − 3 x − 5 y = 5 { 2 x + y = − 3 x − 5 y = 5

ⓐ We will compare the slopes and intercepts of the two lines.

A system of equations whose graphs are parallel lines has no solution and is inconsistent and independent.

ⓑ We will compare the slope and intercepts of the two lines.

A system of equations whose graphs are intersect has 1 solution and is consistent and independent.

Try It 4.11

ⓐ { y = −2 x − 4 4 x + 2 y = 9 { y = −2 x − 4 4 x + 2 y = 9 ⓑ { 3 x + 2 y = 2 2 x + y = 1 { 3 x + 2 y = 2 2 x + y = 1

Try It 4.12

ⓐ { y = 1 3 x − 5 x − 3 y = 6 { y = 1 3 x − 5 x − 3 y = 6 ⓑ { x + 4 y = 12 − x + y = 3 { x + 4 y = 12 − x + y = 3

Solving systems of linear equations by graphing is a good way to visualize the types of solutions that may result. However, there are many cases where solving a system by graphing is inconvenient or imprecise. If the graphs extend beyond the small grid with x and y both between −10 −10 and 10, graphing the lines may be cumbersome. And if the solutions to the system are not integers, it can be hard to read their values precisely from a graph.

Solve a System of Equations by Substitution

We will now solve systems of linear equations by the substitution method.

We will use the same system we used first for graphing.

We will first solve one of the equations for either x or y . We can choose either equation and solve for either variable—but we’ll try to make a choice that will keep the work easy.

Then we substitute that expression into the other equation. The result is an equation with just one variable—and we know how to solve those!

After we find the value of one variable, we will substitute that value into one of the original equations and solve for the other variable. Finally, we check our solution and make sure it makes both equations true.

Example 4.7

How to solve a system of equations by substitution.

Solve the system by substitution: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.13

Solve the system by substitution: { − 2 x + y = −11 x + 3 y = 9 . { − 2 x + y = −11 x + 3 y = 9 .

Try It 4.14

Solve the system by substitution: { 2 x + y = −1 4 x + 3 y = 3 . { 2 x + y = −1 4 x + 3 y = 3 .

Solve a system of equations by substitution.

  • Step 1. Solve one of the equations for either variable.
  • Step 2. Substitute the expression from Step 1 into the other equation.
  • Step 3. Solve the resulting equation.
  • Step 4. Substitute the solution in Step 3 into either of the original equations to find the other variable.
  • Step 5. Write the solution as an ordered pair.
  • Step 6. Check that the ordered pair is a solution to both original equations.

Be very careful with the signs in the next example.

Example 4.8

Solve the system by substitution: { 4 x + 2 y = 4 6 x − y = 8 . { 4 x + 2 y = 4 6 x − y = 8 .

We need to solve one equation for one variable. We will solve the first equation for y .

Try It 4.15

Solve the system by substitution: { x − 4 y = −4 − 3 x + 4 y = 0 . { x − 4 y = −4 − 3 x + 4 y = 0 .

Try It 4.16

Solve the system by substitution: { 4 x − y = 0 2 x − 3 y = 5 . { 4 x − y = 0 2 x − 3 y = 5 .

Solve a System of Equations by Elimination

We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

The Elimination Method is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions a, b, c, and d .

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:

The y ’s add to zero and we have one equation with one variable.

Let’s try another one:

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2 , −2 , we will make the coefficients of x opposites. We must multiply every term on both sides of the equation by −2 . −2 .

Then rewrite the system of equations.

Now we see that the coefficients of the x terms are opposites, so x will be eliminated when we add these two equations.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Example 4.9

How to solve a system of equations by elimination.

Solve the system by elimination: { 2 x + y = 7 x − 2 y = 6 . { 2 x + y = 7 x − 2 y = 6 .

Try It 4.17

Solve the system by elimination: { 3 x + y = 5 2 x − 3 y = 7 . { 3 x + y = 5 2 x − 3 y = 7 .

Try It 4.18

Solve the system by elimination: { 4 x + y = − 5 − 2 x − 2 y = − 2 . { 4 x + y = − 5 − 2 x − 2 y = − 2 .

The steps are listed here for easy reference.

Solve a system of equations by elimination.

  • Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
  • Decide which variable you will eliminate.
  • Multiply one or both equations so that the coefficients of that variable are opposites.
  • Step 3. Add the equations resulting from Step 2 to eliminate one variable.
  • Step 4. Solve for the remaining variable.
  • Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
  • Step 6. Write the solution as an ordered pair.
  • Step 7. Check that the ordered pair is a solution to both original equations.

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Example 4.10

Solve the system by elimination: { 4 x − 3 y = 9 7 x + 2 y = −6 . { 4 x − 3 y = 9 7 x + 2 y = −6 .

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by different constants to get the opposites.

Try It 4.19

Solve the system by elimination: { 3 x − 4 y = − 9 5 x + 3 y = 14 . { 3 x − 4 y = − 9 5 x + 3 y = 14 .

Try It 4.20

Solve each system by elimination: { 7 x + 8 y = 4 3 x − 5 y = 27 . { 7 x + 8 y = 4 3 x − 5 y = 27 .

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by the LCD of all the fractions in the equation.

Example 4.11

Solve the system by elimination: { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 . { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 .

In this example, both equations have fractions. Our first step will be to multiply each equation by the LCD of all the fractions in the equation to clear the fractions.

Try It 4.21

Solve each system by elimination: { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 . { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 .

Try It 4.22

Solve each system by elimination: { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 . { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 .

When we solved the system by graphing, we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

The same is true using substitution or elimination. If the equation at the end of substitution or elimination is a true statement, we have a consistent but dependent system and the system of equations has infinitely many solutions. If the equation at the end of substitution or elimination is a false statement, we have an inconsistent system and the system of equations has no solution.

Example 4.12

Solve the system by elimination: { 3 x + 4 y = 12 y = 3 − 3 4 x . { 3 x + 4 y = 12 y = 3 − 3 4 x .

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 4.23

Solve the system by elimination: { 5 x − 3 y = 15 y = − 5 + 5 3 x . { 5 x − 3 y = 15 y = − 5 + 5 3 x .

Try It 4.24

Solve the system by elimination: { x + 2 y = 6 y = − 1 2 x + 3 . { x + 2 y = 6 y = − 1 2 x + 3 .

Choose the Most Convenient Method to Solve a System of Linear Equations

When you solve a system of linear equations in in an application, you will not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Example 4.13

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

Since both equations are in standard form, using elimination will be most convenient.

Since one equation is already solved for y , using substitution will be most convenient.

Try It 4.25

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = −7 { x = 2 y − 1 3 x − 5 y = −7

Try It 4.26

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

Section 4.1 Exercises

Practice makes perfect.

In the following exercises, determine if the following points are solutions to the given system of equations.

{ 2 x − 6 y = 0 3 x − 4 y = 5 { 2 x − 6 y = 0 3 x − 4 y = 5

ⓐ ( 3 , 1 ) ( 3 , 1 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

{ − 3 x + y = 8 − x + 2 y = −9 { − 3 x + y = 8 − x + 2 y = −9

ⓐ ( −5 , −7 ) ( −5 , −7 ) ⓑ ( −5 , 7 ) ( −5 , 7 )

{ x + y = 2 y = 3 4 x { x + y = 2 y = 3 4 x

ⓐ ( 8 7 , 6 7 ) ( 8 7 , 6 7 ) ⓑ ( 1 , 3 4 ) ( 1 , 3 4 )

{ 2 x + 3 y = 6 y = 2 3 x + 2 { 2 x + 3 y = 6 y = 2 3 x + 2 ⓐ ( −6 , 2 ) ( −6 , 2 ) ⓑ ( −3 , 4 ) ( −3 , 4 )

In the following exercises, solve the following systems of equations by graphing.

{ 3 x + y = −3 2 x + 3 y = 5 { 3 x + y = −3 2 x + 3 y = 5

{ − x + y = 2 2 x + y = −4 { − x + y = 2 2 x + y = −4

{ y = x + 2 y = −2 x + 2 { y = x + 2 y = −2 x + 2

{ y = x − 2 y = −3 x + 2 { y = x − 2 y = −3 x + 2

{ y = 3 2 x + 1 y = − 1 2 x + 5 { y = 3 2 x + 1 y = − 1 2 x + 5

{ y = 2 3 x − 2 y = − 1 3 x − 5 { y = 2 3 x − 2 y = − 1 3 x − 5

{ x + y = −4 − x + 2 y = −2 { x + y = −4 − x + 2 y = −2

{ − x + 3 y = 3 x + 3 y = 3 { − x + 3 y = 3 x + 3 y = 3

{ − 2 x + 3 y = 3 x + 3 y = 12 { − 2 x + 3 y = 3 x + 3 y = 12

{ 2 x − y = 4 2 x + 3 y = 12 { 2 x − y = 4 2 x + 3 y = 12

{ x + 3 y = −6 y = − 4 3 x + 4 { x + 3 y = −6 y = − 4 3 x + 4

{ − x + 2 y = −6 y = − 1 2 x − 1 { − x + 2 y = −6 y = − 1 2 x − 1

{ − 2 x + 4 y = 4 y = 1 2 x { − 2 x + 4 y = 4 y = 1 2 x

{ 3 x + 5 y = 10 y = − 3 5 x + 1 { 3 x + 5 y = 10 y = − 3 5 x + 1

{ 4 x − 3 y = 8 8 x − 6 y = 14 { 4 x − 3 y = 8 8 x − 6 y = 14

{ x + 3 y = 4 − 2 x − 6 y = 3 { x + 3 y = 4 − 2 x − 6 y = 3

{ x = −3 y + 4 2 x + 6 y = 8 { x = −3 y + 4 2 x + 6 y = 8

{ 4 x = 3 y + 7 8 x − 6 y = 14 { 4 x = 3 y + 7 8 x − 6 y = 14

{ 2 x + y = 6 − 8 x − 4 y = −24 { 2 x + y = 6 − 8 x − 4 y = −24

{ 5 x + 2 y = 7 − 10 x − 4 y = −14 { 5 x + 2 y = 7 − 10 x − 4 y = −14

{ y = 2 3 x + 1 − 2 x + 3 y = 5 { y = 2 3 x + 1 − 2 x + 3 y = 5

{ y = 3 2 x + 1 2 x − 3 y = 7 { y = 3 2 x + 1 2 x − 3 y = 7

{ 5 x + 3 y = 4 2 x − 3 y = 5 { 5 x + 3 y = 4 2 x − 3 y = 5

{ y = − 1 2 x + 5 x + 2 y = 10 { y = − 1 2 x + 5 x + 2 y = 10

{ 5 x − 2 y = 10 y = 5 2 x − 5 { 5 x − 2 y = 10 y = 5 2 x − 5

In the following exercises, solve the systems of equations by substitution.

{ 2 x + y = −4 3 x − 2 y = −6 { 2 x + y = −4 3 x − 2 y = −6

{ 2 x + y = −2 3 x − y = 7 { 2 x + y = −2 3 x − y = 7

{ x − 2 y = −5 2 x − 3 y = −4 { x − 2 y = −5 2 x − 3 y = −4

{ x − 3 y = −9 2 x + 5 y = 4 { x − 3 y = −9 2 x + 5 y = 4

{ 5 x − 2 y = −6 y = 3 x + 3 { 5 x − 2 y = −6 y = 3 x + 3

{ − 2 x + 2 y = 6 y = −3 x + 1 { − 2 x + 2 y = 6 y = −3 x + 1

{ 2 x + 5 y = 1 y = 1 3 x − 2 { 2 x + 5 y = 1 y = 1 3 x − 2

{ 3 x + 4 y = 1 y = − 2 5 x + 2 { 3 x + 4 y = 1 y = − 2 5 x + 2

{ 2 x + y = 5 x − 2 y = −15 { 2 x + y = 5 x − 2 y = −15

{ 4 x + y = 10 x − 2 y = −20 { 4 x + y = 10 x − 2 y = −20

{ y = −2 x − 1 y = − 1 3 x + 4 { y = −2 x − 1 y = − 1 3 x + 4

{ y = x − 6 y = − 3 2 x + 4 { y = x − 6 y = − 3 2 x + 4

{ x = 2 y 4 x − 8 y = 0 { x = 2 y 4 x − 8 y = 0

{ 2 x − 16 y = 8 − x − 8 y = −4 { 2 x − 16 y = 8 − x − 8 y = −4

{ y = 7 8 x + 4 − 7 x + 8 y = 6 { y = 7 8 x + 4 − 7 x + 8 y = 6

{ y = − 2 3 x + 5 2 x + 3 y = 11 { y = − 2 3 x + 5 2 x + 3 y = 11

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 − 3 x − y = 0 { 5 x + 2 y = 2 − 3 x − y = 0

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 3 x + 1 2 y = 1 { x + 1 3 y = −1 1 3 x + 1 2 y = 1

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 − 3 x + 12 y = 3 { x − 4 y = −1 − 3 x + 12 y = 3

{ − 3 x − y = 8 6 x + 2 y = −16 { − 3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Writing Exercises

In a system of linear equations, the two equations have the same intercepts. Describe the possible solutions to the system.

Solve the system of equations by substitution and explain all your steps in words: { 3 x + y = 12 x = y − 8 . { 3 x + y = 12 x = y − 8 .

Solve the system of equations by elimination and explain all your steps in words: { 5 x + 4 y = 10 2 x = 3 y + 27 . { 5 x + 4 y = 10 2 x = 3 y + 27 .

Solve the system of equations { x + y = 10 x − y = 6 { x + y = 10 x − y = 6

ⓐ by graphing ⓑ by substitution ⓒ Which method do you prefer? Why?

After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

If most of your checks were:

…confidently. Congratulations! You have achieved the objectives in this section. Reflect on the study skills you used so that you can continue to use them. What did you do to become confident of your ability to do these things? Be specific.

…with some help. This must be addressed quickly because topics you do not master become potholes in your road to success. In math every topic builds upon previous work. It is important to make sure you have a strong foundation before you move on. Whom can you ask for help?Your fellow classmates and instructor are good resources. Is there a place on campus where math tutors are available? Can your study skills be improved?

…no - I don’t get it! This is a warning sign and you must not ignore it. You should get help right away or you will quickly be overwhelmed. See your instructor as soon as you can to discuss your situation. Together you can come up with a plan to get you the help you need.

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Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
  • Authors: Lynn Marecek, Andrea Honeycutt Mathis
  • Publisher/website: OpenStax
  • Book title: Intermediate Algebra 2e
  • Publication date: May 6, 2020
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
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Module 5: Systems of Linear Equations

5.2 – applications of systems of linear equations, learning objectives.

  • Specify what the variables in a cost/ revenue system of linear equations represent
  • Determine and apply an appropriate method for solving the system

(5.2.2) – Solve value problems with a system of linear equations

(5.2.3) – solve mixture problems with a system of linear equations, (5.2.4) – solve uniform motion problems with a system of linear equations.

A skateboard manufacturer introduces a new line of boards. The manufacturer tracks its costs, which is the amount it spends to produce the boards, and its revenue, which is the amount it earns through sales of its boards. How can the company determine if it is making a profit with its new line? How many skateboards must be produced and sold before a profit is possible?

Skateboarders at a skating rink by the beach.

(credit: Thomas Sørenes)

(5.2.1) – Solve cost and revenue problems

Using what we have learned about systems of equations, we can return to the skateboard manufacturing problem at the beginning of the section. The skateboard manufacturer’s revenue function is the function used to calculate the amount of money that comes into the business. It can be represented by the equation [latex]R=xp[/latex], where [latex]x=[/latex] quantity and [latex]p=[/latex] price. The revenue function is shown in orange in the graph below.

The cost function is the function used to calculate the costs of doing business. It includes fixed costs, such as rent and salaries, and variable costs, such as utilities. The cost function is shown in blue in the graph below. The [latex]x[/latex] -axis represents quantity in hundreds of units. The y -axis represents either cost or revenue in hundreds of dollars.

A graph showing money in hundreds of dollars on the y axis and quantity in hundreds of units on the x axis. A line representing cost and a line representing revenue cross at the point (7,33), which is marked break-even. The shaded space between the two lines to the right of the break-even point is labeled profit.

The point at which the two lines intersect is called the break-even point . We can see from the graph that if 700 units are produced, the cost is $3,300 and the revenue is also $3,300. In other words, the company breaks even if they produce and sell 700 units. They neither make money nor lose money.

The shaded region to the right of the break-even point represents quantities for which the company makes a profit. The shaded region to the left represents quantities for which the company suffers a loss. The profit function is the revenue function minus the cost function, written as [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex]. Clearly, knowing the quantity for which the cost equals the revenue is of great importance to businesses.

A business wants to manufacture bike frames. Before they start production, they need to make sure they can make a profit with the materials and labor force they have. Their accountant has given them a cost equation of [latex]y=0.85x+35,000[/latex] and a revenue equation of [latex]y=1.55x[/latex]:

  • Interpret x and y for the cost equation
  • Interpret x and y for the revenue equation

Cost: [latex]y=0.85x+35,000[/latex]

Revenue:[latex]y=1.55x[/latex]

The cost equation represents money leaving the company, namely how much it costs to produce a given number of bike frames. If we use the skateboard example as a model, x would represent the number of frames produced (instead of skateboards) and y would represent the amount of money it would cost to produce them (the same as the skateboard problem).

The revenue equation represents money coming into the company, so in this context x still represents the number of bike frames manufactured, and y now represents the amount of money made from selling them.  Let’s organize this information in a table:

Example: Finding the Break-Even Point and the Profit Function Using Substitution

Given the cost function [latex]C\left(x\right)=0.85x+35,000[/latex] and the revenue function [latex]R\left(x\right)=1.55x[/latex], find the break-even point and the profit function.

Write the system of equations using [latex]y[/latex] to replace function notation.

[latex]\begin{array}{l}\begin{array}{l}\\ y=0.85x+35,000\end{array}\hfill \\ y=1.55x\hfill \end{array}[/latex]

Substitute the expression [latex]0.85x+35,000[/latex] from the first equation into the second equation and solve for [latex]x[/latex].

[latex]\begin{array}{c}0.85x+35,000=1.55x\\ 35,000=0.7x\\ 50,000=x\end{array}[/latex]

Then, we substitute [latex]x=50,000[/latex] into either the cost function or the revenue function. [latex]1.55\left(50,000\right)=77,500[/latex]

The break-even point is [latex]\left(50,000,77,500\right)[/latex].

The profit function is found using the formula [latex]P\left(x\right)=R\left(x\right)-C\left(x\right)[/latex].

[latex]\begin{array}{l}P\left(x\right)=1.55x-\left(0.85x+35,000\right)\hfill \\ \text{ }=0.7x - 35,000\hfill \end{array}[/latex]

The profit function is [latex]P\left(x\right)=0.7x - 35,000[/latex].

Analysis of the Solution

The cost to produce 50,000 units is $77,500, and the revenue from the sales of 50,000 units is also $77,500. To make a profit, the business must produce and sell more than 50,000 units.

A graph showing money in dollars on the y axis and quantity on the x axis. A line representing cost and a line representing revenue cross at the break-even point of fifty thousand, seventy-seven thousand five hundred. The cost line's equation is C(x)=0.85x+35,000. The revenue line's equation is R(x)=1.55x. The shaded space between the two lines to the right of the break-even point is labeled profit.

We see from the graph below that the profit function has a negative value until [latex]x=50,000[/latex], when the graph crosses the x -axis. Then, the graph emerges into positive y -values and continues on this path as the profit function is a straight line. This illustrates that the break-even point for businesses occurs when the profit function is 0. The area to the left of the break-even point represents operating at a loss.

A graph showing dollars profit on the y axis and quantity on the x axis. The profit line crosses the break-even point at fifty thousand, zero. The profit line's equation is P(x)=0.7x-35,000.

It is rare to be given equations that neatly model behaviors that you encounter in business, rather, you will probably be faced with a situation for which you know key information as in the example above. Below, we summarize three key factors that will help guide you in translating a situation into a system.

How To: Given a situation that represents a system of linear equations, write the system of equations and identify the solution.

1) Identify unknown quantities in a problem represent them with variables.

2) Write a system of equations which models the problem’s conditions.

3) Solve the system.

4) Check proposed solution.

Now let’s practice putting these key factors to work. In the next example, we determine how many different types of tickets are sold given information about the total revenue and amount of tickets sold to an event.

Example: Writing and Solving a System of Equations in Two Variables

The cost of a ticket to the circus is $25.00 for children and $50.00 for adults. On a certain day, attendance at the circus is 2,000 and the total gate revenue is $70,000. How many children and how many adults bought tickets?

Let c = the number of children and a = the number of adults in attendance.

The total number of people is [latex]2,000[/latex]. We can use this to write an equation for the number of people at the circus that day.

[latex]c+a=2,000[/latex]

The revenue from all children can be found by multiplying $25.00 by the number of children, [latex]25c[/latex]. The revenue from all adults can be found by multiplying $50.00 by the number of adults, [latex]50a[/latex]. The total revenue is $70,000. We can use this to write an equation for the revenue.

[latex]25c+50a=70,000[/latex]

We now have a system of linear equations in two variables.

[latex]\begin{array}{c}c+a=2,000\\ 25c+50a=70,000\end{array}[/latex]

In the first equation, the coefficient of both variables is 1. We can quickly solve the first equation for either [latex]c[/latex] or [latex]a[/latex]. We will solve for [latex]a[/latex].

[latex]\begin{array}{c}c+a=2,000\\ a=2,000-c\end{array}[/latex]

Substitute the expression [latex]2,000-c[/latex] in the second equation for [latex]a[/latex] and solve for [latex]c[/latex].

[latex]\begin{array}{l} 25c+50\left(2,000-c\right)=70,000\hfill \\ 25c+100,000 - 50c=70,000\hfill \\ \text{ }-25c=-30,000\hfill \\ \text{ }c=1,200\hfill \end{array}[/latex]

Substitute [latex]c=1,200[/latex] into the first equation to solve for [latex]a[/latex].

[latex]\begin{array}{l}1,200+a=2,000\hfill \\ \text{ }\text{}a=800\hfill \end{array}[/latex]

We find that [latex]1,200[/latex] children and [latex]800[/latex] adults bought tickets to the circus that day.

In this video example we show how to set up a system of linear equations that represents the total cost for admission to a museum.

Meal tickets at the circus cost $4.00 for children and $12.00 for adults. If 1,650 meal tickets were bought for a total of $14,200, how many children and how many adults bought meal tickets?

700 children, 950 adults

Sometimes, a system can inform a decision.  In our next example, we help answer the question, “Which truck rental company will give the best value?”

Example: Building a System of Linear Models to Choose a Truck Rental Company

Jamal is choosing between two truck-rental companies. The first, Keep on Trucking, Inc., charges an up-front fee of $20, then 59 cents a mile. The second, Move It Your Way, charges an up-front fee of $16, then 63 cents a mile. [1] When will Keep on Trucking, Inc. be the better choice for Jamal?

The two important quantities in this problem are the cost and the number of miles driven. Because we have two companies to consider, we will define two functions.

A linear function is of the form [latex]f\left(x\right)=mx+b[/latex]. Using the rates of change and initial charges, we can write the equations

[latex]\begin{array}{l}K\left(d\right)=0.59d+20\\ M\left(d\right)=0.63d+16\end{array}[/latex]

Using these equations, we can determine when Keep on Trucking, Inc., will be the better choice. Because all we have to make that decision from is the costs, we are looking for when Move It Your Way, will cost less, or when [latex]K\left(d\right)<M\left(d\right)[/latex]. The solution pathway will lead us to find the equations for the two functions, find the intersection, and then see where the [latex]K\left(d\right)[/latex] function is smaller.

image

These graphs are sketched above, with K ( d ) in blue.

To find the intersection, we set the equations equal and solve:

[latex]\begin{array}{l}K\left(d\right)=M\left(d\right)\hfill \\ 0.59d+20=0.63d+16\hfill \\ 4=0.04d\hfill \\ 100=d\hfill \\ d=100\hfill \end{array}[/latex]

This tells us that the cost from the two companies will be the same if 100 miles are driven. Either by looking at the graph, or noting that [latex]K\left(d\right)[/latex] is growing at a slower rate, we can conclude that Keep on Trucking, Inc. will be the cheaper price when more than 100 miles are driven, that is [latex]d>100[/latex].

One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution.  A solution is a mixture of two or more different substances like water and salt or vinegar and oil.  Most biochemical reactions occur in liquid solutions, making them important for doctors, nurses, and researchers to understand.  There are many other disciplines that use solutions as well.

The concentration or strength of a liquid solution is often described  as a percentage.  This number comes from the ratio of how much mass is in a specific volume of liquid.  For example if you have 50 grams of salt in a 100mL of water you have a 50% salt solution based on the following ratio:

[latex]\frac{50\text{ grams }}{100\text{ mL }}=0.50\frac{\text{ grams }}{\text{ mL }}=50\text{ % }[/latex]

Solutions used for most purposes typically come in pre-made concentrations from manufacturers, so if you need a custom concentration, you would need to mix two different strengths.  In this section, we will practice writing equations that represent the outcome from mixing two different concentrations of solutions.

We will use the following table to help us solve mixture problems:

To demonstrate why the table is helpful in solving for unknown amounts or concentrations of a solution, consider two solutions that are mixed together, one is 120mL of a 9% solution, and the other is 75mL of a 23% solution. If we mix both of these solutions together we will have a new volume and a new mass of solute and with those we can find a new concentration.

First, find the total mass of solids for each solution by multiplying the volume by the concentration.

Next we add the new volumes and new masses.

Now we have used mathematical operations to describe the result of mixing two different solutions. We know the new volume, concentration and mass of solute in the new solution.  In the following examples, you will see that we can use the table to find an unknown final volume or concentration. These problems can have either one or two variables. We will start with one variable problems, then move to two variable problems.

A chemist has 70 mL of a 50% methane solution. How much of an 80% solution must she add so the final solution is 60% methane?

Let’s use the problem solving process outlined in Module 1 to help us work through a solution to the problem.

Read and Understand:  We are looking for a new amount – in this case a volume –  based on the words “how much”.  We know two starting  concentrations and the final concentration, as well as one volume.

Define and Translate:  Solution 1 is the 70 mL of 50% methane and solution 2 is the unknown amount with 80% methane.  We can call our unknown amount x.

Write and Solve:   Set up the mixture table. Remember that concentrations are written as decimals before we can perform mathematical operations on them.

Multiply amount by concentration to get total, be sure to distribute on the last row: [latex]\left(70 + x\right)0.6[/latex]Add the entries in the amount column to get final amount. The concentration for this amount is 0.6 because we want the final solution to be 60% methane.

Add the total mass for solution 1 and solution 2 to get the total mass for the 60% solution. This is our equation for finding the unknown volume.

[latex]35+0.8x=42+0.6x[/latex]

[latex]\begin{array}{c}35+0.8x=42+0.6x\\\underline{-0.6x}\,\,\,\,\,\,\,\underline{-0.6x}\\35+0.2x=42\\\end{array}[/latex]

Subtract 35 from both sides

[latex]\begin{array}{c}35+0.2x=42\\\underline{-35}\,\,\,\,\,\,\,\underline{-35}\\0.2x=7\end{array}[/latex]

Divide both sides by 0.2

[latex]\begin{array}{c}0.2x=7\\\frac{0.2x}{0.2}=\frac{7}{0.2}\end{array}[/latex] [latex]x=35[/latex]

35mL must be added to the original 70 mL to gain a solution with a concentration of 60%

The above problem illustrates how we can use the mixture table to define an equation to solve for an unknown volume. In the next example we will start with two known concentrations and use a system of equations to find two starting volumes necessary to achieve a specified final concentration.

A farmer has two types of milk, one that is 24% butterfat and another which is 18% butterfat. How much of each should he use to end up with 42 gallons of 20% butterfat?

Read and Understand:  We are asked to find two starting volumes of milk whose concentrations of butterfat are both known. We also know the final volume is 42 gallons. There are two unknowns in this problem.

Define and Translate:  We will call the unknown volume of the  24% solution x, and the unknown volume of the 18% solution y.

Write and Solve:  Fill in the table with the information we know.

Find the total mass by multiplying the amount of each solution by the concentration. The total mass of the final solution comes from

When you sum the amount column you get one equation: [latex]x+ y = 42[/latex] When you sum the total column you get a second equation: [latex]0.24x + 0.18y = 8.4[/latex]

Use elimination to find a value for [latex]x[/latex], and [latex]y[/latex].

Multiply the first equation by [latex]-0.18[/latex]

[latex]\begin{array}{cc}-0.18(x+y) &= (42)(-0.18) \\ -0.18x-0.18y &= -7.56 \end{array}[/latex]

Now our system of equations looks like this:

[latex]\begin{array}{cc} -0.18x-0.18y &= -7.56\\0.24x + 0.18y &= 8.4 \end{array}[/latex]

Adding the two equations together to eliminate the y terms gives this equation:

[latex]0.06x = 8.4[/latex]

Divide by 0.06 on each side:

[latex]x = 14[/latex]

Now substitute the value for x into one of the equations in order to solve for y.

[latex]\begin{array}{cc} (14) + y &= 42\\ y &= 28 \end{array}[/latex]

This can be interpreted as 14 gallons of 24% butterfat milk added to 28 gallons of 18% butterfat milk will give 42 gallons of 20% butterfat milk.

In the following video you will be given an example of how to solve a mixture problem without using a table, and interpret the results.

Many real-world applications of uniform motion arise because of the effects of currents—of water or air—on the actual speed of a vehicle. Cross-country airplane flights in the United States generally take longer going west than going east because of the prevailing wind currents.

Let’s take a look at a boat travelling on a river. Depending on which way the boat is going, the current of the water is either slowing it down or speeding it up.

The images below show how a river current affects the speed at which a boat is actually travelling. We’ll call the speed of the boat in still water [latex]b[/latex] and the speed of the river current [latex]c[/latex].

The boat is going downstream, in the same direction as the river current. The current helps push the boat, so the boat’s actual speed is faster than its speed in still water. The actual speed at which the boat is moving is [latex]b+c[/latex].

using systems of two equations to solve application problems

Now, the boat is going upstream, opposite to the river current. The current is going against the boat, so the boat’s actual speed is slower than its speed in still water. The actual speed of the boat is [latex]b-c[/latex].

using systems of two equations to solve application problems

We’ll put some numbers to this situation in the next example.

Translate to a system of equations and then solve.

A river cruise ship sailed 60 miles downstream for 4 hours and then took 5 hours sailing upstream to return to the dock. Find the speed of the ship in still water and the speed of the river current.

Read the problem: This is a uniform motion problem and a picture will help us visualize the situation.

using systems of two equations to solve application problems

Identify  what we are looking for: We are looking for the speed of the ship in still water and the speed of the current.

Name  what we are looking for:

Let [latex]s=[/latex] the rate of the ship in still water.

Let [latex]c=[/latex] the rate of the current.

A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship and so the ship’s actual rate is [latex]s+c[/latex]. Going upstream, the current slows the ship and so the actual rate is [latex]s-c[/latex]. Downstream it takes 4 hours. Upstream it takes 5 hours. Each way the distance is 60 miles.

using systems of two equations to solve application problems

Translate  into a system of equations. Since rate times time is distance, we can write the system of equations.

 [latex]\begin{array}{c}4(s+c)=60 \\ 5(s-c) = 60\end{array}[/latex]

Solve  the system of equations. Distribute to put both equations in standard form, then solve by elimination.

 [latex]\begin{array}{c}4s+4c=60 \\ 5s-5c = 60\end{array}[/latex]

Multiply the top equation by 5 and the bottom equation by 4. Add the equations, then solve for [latex]s[/latex].

[latex]\begin{array}{cc}20s+20c &= 300 \\ 20s-20c &= 240 \\ \hline \\ 40s &= 540 \\ s &= 13.5 \end{array}[/latex]

Substitute [latex]s=13.5[/latex] into one of the original equations.

[latex]\begin{array}{cc}4(s+c) &= 60 \\ 4(13.5+c) &= 60 \\ 54 + 4c &= 60 \\ 4c &= 6 \\ c &= 1.5 \end{array}[/latex]

Check  the answer in the problem. The downstream rate would be:

[latex]13.5+1.5 = 15[/latex] mph

In 4 hours the ship would travel:

[latex]15 \cdot 4 = 60[/latex] miles.

The upstream rate would be

[latex]13.5 - 1.5 = 12[/latex] mph.

In 5 hours the ship would travel

[latex]12\cdot 5[/latex] miles.

Answer  the question. The rate of the ship is 13.5 mph and the rate of the current is 1.5 mph.

In the next video, we present another example of a uniform motion problem which can be solved with a system of linear equations.

  • Rates retrieved Aug 2, 2010 from http://www.budgettruck.com and http://www.uhaul.com/ ↵
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  • College Algebra. Authored by : Abramson, Jay et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
  • Solving Systems of Equations using Elimination. Authored by : James Sousa (Mathispower4u.com). Located at : https://youtu.be/ova8GSmPV4o . License : CC BY: Attribution
  • Question ID 115164, 115120, 115110. Authored by : Shabazian, Roy. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
  • Beginning and Intermediate Algebra. Authored by : Wallace, Tyler. Located at : http://www.wallace.ccfaculty.org/book/book.html . License : CC BY: Attribution
  • Question ID 29699. Authored by : McClure, Caren. License : CC BY: Attribution . License Terms : IMathAS Community License CC-BY + GPL
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  • Ex: System of Equations Application - Mixture Problem.. Authored by : James Sousa (Mathispower4u.com) for Lumen Learning.. Located at : https://youtu.be/4s5MCqphpKo. . License : CC BY: Attribution
  • Beginning and Intermediate Algebra Textbook. . Authored by : Tyler Wallace. Located at : . License : CC BY: Attribution
  • Ex: System of Equations Application - Plane and Wind problem. Authored by : James Sousa (Mathispower4u.com). Located at : https://www.youtube.com/watch?v=OuxMYTqDhxw . License : CC BY: Attribution
  • Intermediate Algebra . Authored by : Lynn Marecek et al.. Provided by : OpenStax. Located at : http://cnx.org/contents/[email protected] . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]
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Applications of Systems of Equations

Related Topics: More Lessons for Basic Algebra Math Worksheets

A series of free Basic Algebra Lessons.

In this lesson, we will learn

  • mixture word problems
  • rate word problems
  • work word problems
  • money word problems

Mixture Problems

Some word problems using systems of equations involve mixing two quantities with different prices. To solve mixture problems, knowledge of solving systems of equations. is necessary. Most often, these problems will have two variables, but more advanced problems have systems of equations with three variables. Other types of word problems using systems of equations include rate word problems and work word problems.

Explains the concept of a value mixture problem and works this problem. Example: Two pounds of organic tea that costs $6.75 a pound is mixed with some generic tea that costs $3.25 a pound. How many pounds of the generic tea should be used to make a new tea mixture that costs $4.65 a pound?

Mixture Problem Example: A special treat is made by mixing 5 pounds of popcorn costing $.80 a pound with caramel costing $2.40 a pound. How many pounds of caramel are needed to make a mixture that costs $1.40 a pound?

Mixture Problem Example: Tina wants to mix some blackberries that cost $2.75 a pound with 2 pounds of raspberries that cost $3.50 a pound to get a mixture that costs $3.25 a pound. How many blackberries should Tina buy?

Rate Word Problems

To solve rate word problems, knowledge of solving systems of equations is necessary. Rate word problems include problems dealing with rates, distances, time and wind or water current. Other types of word problems using systems of equations include money word problems and age word problems. Solves this word problem using uniform motion* rt = d* formula Example: Two cyclists start at the same corner and ride in opposite directions. One cyclist rides twice as fast as the other. In 3 hours, they are 81 miles apart. Find the rate of each cyclist.

Solves this word problem using uniform motion rt = d formula Example: A 555-mile, 5-hour trip on the Autobahn was driven at two speeds. The average speed of the car was 105 mph on the first part of the trip, and the average speed was 115 mph for the second part. How long did the car drive at each speed?

Algebra Word Problem: Distance, Rate, and Time. This video shows an example about finding the speed of the current of stream.

Work Word Problems

It is possible to solve word problems when two people are doing a work job together by solving systems of equations. To solve a work word problem, multiply the hourly rate of the two people working together times the time spent working to get the total amount of time spent on the job. Knowledge of solving systems of equations is necessary to solve these types of problems. Linear Equation , Word Problem: Work, Rates, Time To Complete a Task. Example: Given that a person can complete a task alone in 32 hours and with another person they can finish the task in 19 hours. How long it would take the second person working alone?

Linear Equation , Word Problem : Work, Rates, Time To Complete a Task. Example: A drain is emptying a pool while a hose is filling the pool at the same time. Find how long it would take to fill the empty pool.

Money Word Problems

Setting up a system of equations is usually effective when solving money word problems and word problems involving coins. A knowledge of how to solve systems of equations is necessary to solve these types of equations. Other types of word problems using systems of equations include mixture problems, rate word problems, work word problems and age word problems. Algebra Money Word Problems with two variables (x and y). Solving equations with two variables

Algebra word problem involving money

Word problem involving coins and money

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Physics LibreTexts

2.9: Solving a System of Linear Equations

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Learning Objectives

  • Solve a system of equations using the substitution method.
  • Recognize systems of equations that have no solution or an infinite number of solutions.
  • Solve application problems using the substitution method.
  • Solve a system of equations when no multiplication is necessary to eliminate a variable.
  • Solve a system of equations when multiplication is necessary to eliminate a variable.
  • Recognize systems that have no solution or an infinite number of solutions.
  • Solve application problems using the elimination method.
  • Solve application problems that require the use of this method.

Using Substitution to Solve a System of Equations

In the substitution method, you solve for one variable and then substitute that expression into the other equation. The important thing here is that you are always substituting values that are equivalent.

For example:

Sean is 5 years older than four times his daughter’s age. His daughter is 7. How old is Sean?

You might do this problem in your head. Sean’s daughter is 7, so “four times his daughter’s age” is 28, and 5 years added to that is 33. Sean is 33.

If you solved the problem like that, you used a simple substitution—you substituted in the value “7” for “his daughter’s age.” You learned in the second part of the problem that “his daughter is 7.” So substituting in a value of “7” for “his daughter’s age” in the first part of the problem was okay, because you knew these two quantities were equal.

Let’s look at a simple system of equations that can be solved using substitution.

Find the value of \(\ x\) for this system.

Equation A: \(\ 4 x+3 y=-14\)

Equation B: \(\ y=2\)

You can substitute a value for a variable even if it is an expression. Here’s an example.

Solve for \(\ x\) and \(\ y\).

Equation A: \(\ y+x=3\)

Equation B: \(\ x=y+5\)

Remember, a solution to a system of equations must be a solution to each of the equations within the system. The ordered pair \(\ (4,-1)\) works for both equations, so you know that it is a solution to the system as well.

Let’s look at another example whose substitution involves the distributive property.

\(\ \begin{array}{c} y=3 x+6 \\ -2 x+4 y=4 \end{array}\)

\(\ x=-2\) and \(\ y=0\).

The solution is \(\ (-2,0)\).

In the examples above, one of the equations was already given to us in terms of the variable \(\ x\) or \(\ y\). This allowed us to quickly substitute that value into the other equation and solve for one of the unknowns.

Sometimes you may have to rewrite one of the equations in terms of one of the variables first before you can substitute. Look at the example below.

\(\ \begin{array}{c} 2 x+3 y=22 \\ 3 x+y=19 \end{array}\)

\(\ x=5\) and \(\ y=4\)

The solution is \(\ (5,4)\).

Special Situations

There are some cases where using the substitution method will yield results that, at first, do not make sense. Let’s take a look at some of these and figure out what is going on.

\(\ \begin{array}{c} y=5 x+4 \\ 10 x-2 y=4 \end{array}\)

The statement \(\ -8=4\) is false, so there is no solution.

You get the false statement \(\ -8=4\). What does this mean? The graph of this system sheds some light on what is happening.

Screen Shot 2021-06-29 at 4.29.57 PM.png

The lines are parallel. They never intersect and there is no solution to this system of linear equations. Note that the result \(\ -8=4\) is not a solution. It is simply a false statement and it indicates that there is no solution.

Now take this problem, which is interesting as well.

\(\ \begin{array}{c} \color{green}\text { Solve for } x \text { and } y.\\ \color{green}y=-0.5x\\ \color{green}9y=-4.5x \end{array}\)

Substituting \(\ -0.5 x\) for \(\ y\) in the second equation, you find the following:

\(\ \begin{aligned} 9 y &=-4.5 x \\ 9(-0.5 x) &=-4.5 x \\ -4.5 x &=-4.5 x \end{aligned}\)

This time, you get a true statement: \(\ -4.5 x=-4.5 x\). But what does this type of answer mean? Again, graphing can help you make sense of this system.

Screen Shot 2021-06-29 at 8.11.27 PM.png

This system consists of two equations that both represent the same line; the two lines are collinear. Every point along the line will be a solution to the system, and that’s why the substitution method yields a true statement. In this case, there are an infinite number of solutions.

Solving Application Problems Using Substitution

Systems of equations are a very useful tool for modeling real-life situations and answering questions about them. If you can translate the application into two linear equations with two variables, then you have a system of equations that you can solve to find the solution. You can use any method to solve the system of equations. Use the substitution method in this topic.

In order to sell more of its produce, a local farm sells bags of apples in two sizes: medium and large. A medium bag contains 4 Macintosh and 1 Granny Smith apples and costs $2.80. A large bag contains 8 Macintosh and 4 Granny Smith apples and costs $7.20. The price of one Granny Smith apple is the same in the medium bag as it is in the large bag. The price of one Macintosh apple is the same in the medium bag as it is in the large bag. What is the price of each kind of apple?

Let’s start by creating a system of equations that represents what is happening in the problem. There are two types of apples and two sizes of bags. You can let \(\ m\) represent the cost of a Macintosh apple and \(\ g\) represent the cost of a Granny Smith apple. Let’s make a table and see what is known.

Now that you have two equations in the same variables, you can solve the system. You will use substitution. The steps are shown in the example below:

Solve for \(\ g\) and \(\ m\) using the substitution method.

\(\ \begin{array}{c} 4 m+g=2.80 \\ 8 m+4 g=7.20 \end{array}\)

One Granny Smith apple costs $0.80 and one Macintosh apple costs $0.50.

Using the substitution method can be an effective approach to solving geometric problems.

The perimeter of a rectangle is 60 inches. If the length is 10 inches longer than the width, find the dimensions using the substitution method.

The length of the rectangle is 20 inches.

The width of the rectangle is 10 inches.

Solving a System of Equations Using the Elimination Method

The elimination method for solving systems of linear equations uses the addition property of equality. You can add the same value to each side of an equation.

So if you have a system: \(\ x-y=-6\) and \(\ x+y=8\), you can add \(\ x+y\) to the left side of the first equation and add 8 to the right side of the equation. And since \(\ x+y=8\), you are adding the same value to each side of the first equation.

Using Addition to Eliminate a Variable

If you add the two equations, \(\ x-y=-6\) and \(\ x+y=8\) together, as noted above, watch what happens.

\(\ \begin{array}{rr} x-y= & -6 \\ x+y= & 8 \\ \hline 2 x+0= & 2 \end{array}\)

You have eliminated the \(\ y\) term, and this equation can be solved using the methods for solving equations with one variable.

Let’s see how this system is solved using the elimination method.

Use elimination to solve the system.

\(\ \begin{array}{c} x-y=-6 \\ x+y=8 \end{array}\)

Be sure to check your answer in both equations!

The answers check.

The solution is \(\ (1,7)\).

Unfortunately, not all systems work out this easily. How about a system like \(\ 2 x+y=12\) and \(\ -3 x+y=2\)? If you add these two equations together, no variables are eliminated.

\(\ \begin{aligned} 2 x+y&=12 \\ -3 x+y&=2 \\ \hline-x+2 y&=14 \end{aligned}\)

But you want to eliminate a variable. So let’s add the opposite of one of the equations to the other equation.

\(\ \begin{array}{r} 2 x+y=12 \quad \rightarrow& 2 x+y=12\quad \rightarrow& \quad 2 x+y=\ 12 \\ -3 x+y=2 \quad\rightarrow&-(-3 x+y)=-(2)\quad \rightarrow& \quad \underline{3x-y=-2}\\ && 5x+0y=\ 10 \end{array}\)

You have eliminated the \(\ y\) variable, and the problem can now be solved. See the example below.

\(\ \begin{aligned} 2 x+y&=12 \\ -3 x+y&=2 \end{aligned}\)

The solution is \(\ (2,8)\).

The following are two more examples showing how to solve linear systems of equations using elimination.

\(\ \begin{aligned} -2 x+3 y&=-1 \\ 2 x+5 y&=25 \end{aligned}\)

Check solutions.

The solution is \(\ (5,3)(5,3)\).

Use elimination to solve for \(\ x\) and \(\ y\).

The solution is \(\ (2,3)\).

Go ahead and check this last example—substitute \(\ (2,3)\) into both equations. You get two true statements: \(\ 14=14\) and \(\ 16=16\)!

Notice that you could have used the opposite of the first equation rather than the second equation and gotten the same result.

Using Multiplication and Addition to Eliminate a Variables

Many times, adding the equations or adding the opposite of one of the equations will not result in eliminating a variable. Look at the system below.

\(\ \begin{aligned} 3 x+4 y=52 \\ 5 x+y=30 \end{aligned}\)

If you add the equations above, or add the opposite of one of the equations, you will get an equation that still has two variables. So let’s now use the multiplication property of equality first. You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation.

This is where multiplication comes in handy. The first equation contains the term \(\ 4y\) and the second equation contains the term \(\ y\). If you multiply the second equation by -4, when you add both equations, the \(\ y\) variables will add up to 0.

\(\ \begin{array}{l} 3 x+4 y=52 \quad&\rightarrow \quad3 x+4 y=52 \quad&\rightarrow \quad3 x+4 y= 52 \\ 5 x+y=30 \quad&\rightarrow \quad-4(5 x+y)=-4(30) \quad&\rightarrow \quad\underline{-20 x-4 y=-120}\\ &&\quad-17x+0y=-68 \end{array}\)

See the example below.

Equation A: \(\ 3 x+4 y=52\)

Equation B: \(\ 5 x+y=30\)

Check your answer.

The solution is \(\ (4,10)\).

There are other ways to solve this system. Instead of multiplying one equation in order to eliminate a variable when the equations were added, you could have multiplied both equations by different numbers.

Let’s remove the variable \(\ x\) this time. Multiply Equation A by 5 and Equation B by -3.

\(\ \begin{array}{r} 3 x+4 y=52 \\ 5 x+\ \ y=30 \end{array}\)

These equations were multiplied by 5 and -3 respectively, because that gave you terms that would add up to 0. Be sure to multiply all of the terms of the equation.

Just as with the substitution method, the elimination method will sometimes eliminate both variables, and you end up with either a true statement or a false statement. Recall that a false statement means that there is no solution.

Let’s look at an example.

\(\ \begin{array}{c} -x-y=-4 \\ x+y=2 \end{array}\)

There is no solution.

Graphing these lines shows that they are parallel lines and as such do not share any point in common, verifying that there is no solution.

Screen Shot 2021-06-29 at 11.40.42 PM.png

If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. In fact, the equations are the same line.

\(\ \begin{array}{r} x+y&=&2 \\ -x-y&=&-2 \end{array}\)

There are an infinite number of solutions.

Graphing these two equations will help to illustrate what is happening.

Screen Shot 2021-06-29 at 11.47.25 PM.png

Solving Application Problems Using the Elimination Method

The elimination method can be applied to solving systems of equations that model real situations. Two examples of using the elimination method in problem solving are shown below.

The sum of two numbers is 10. Their difference is 6. What are the two numbers?

Check your answer by substituting \(\ x=8\) and \(\ y=2\) into the original system.

The numbers are 8 and 2.

A theater sold 800 tickets for Friday night’s performance. One child ticket costs $4.50 and one adult ticket costs $6.00. The total amount collected was $4,500. How many of each type of ticket were sold?

Check your answer by substituting \(\ a=600\) and \(\ c=200\) into the original system.

600 adult tickets and 200 child tickets were sold.

Solving A System of Three Linear Equations with Three Variables

Equations can have more than one or two variables. You are going to look at equations with three variables. Equations with one variable graph on a line. Equations with two variables graph on a plane. Equations with three variables graph in a 3-dimensional space.

Equations with one variable require only one equation to have a unique (one) solution. Equations with two variables require two equations to have a unique solution (one ordered pair). So it should not be a surprise that equations with three variables require a system of three equations to have a unique solution (one ordered triplet).

Solving A System of Three Variables

Just as when solving a system of two equations, there are three possible outcomes for the solution of a system of three variables. Let’s look at this visually, although you will not be graphing these equations.

Case 1: There is one solution. In order for three equations with three variables to have one solution, the planes must intersect in a single point.

Screen Shot 2021-06-30 at 10.02.53 AM.png

Case 2: There is no solution. The three planes do not have any points in common. (Note that two of the equations may have points in common with each other, but not all three.) Below are examples of some of the ways this can happen.

Screen Shot 2021-06-30 at 10.03.57 AM.png

Case 3: There are an infinite number of solutions. This occurs when the three planes intersect in a line. And this can also occur when the three equations graph as the same plane.

Screen Shot 2021-06-30 at 10.09.07 AM.png

Let’s start by looking at Case 1, where the system has a unique (one) solution. This is the case that you are usually most interested in.

Here is a system of linear equations. There are three variables and three equations.

\(\ \begin{array}{l} 3 x+4 y-z=8 \\ 5 x-2 y+z=4 \\ 2 x-2 y+z=1 \end{array}\)

You know how to solve a system with two equations and two variables. For the first step, use the elimination method to remove one of the variables. In this case, \(\ z\) can be eliminated by adding the first and second equations.

\(\ \begin{aligned} 3 x+4 y-z&=\ \ 8 \\ 5 x-2 y+z&=\ \ 4 \\ \hline 8 x+2 y\ \ \ \ \ \ \ &=12 \end{aligned}\)

To solve the system, though, you need two equations using two variables. Adding the first and third equations in the original system will also give an equation with \(\ x\) and \(\ y\) but not \(\ z\).

\(\ \begin{array}{r} 3 x+4 y-z=8 \\ 2 x-2 y+z=1 \\ \hline 5 x+2 y\ \ \ \quad=9 \end{array}\)

Now you have a system of two equations and two variables.

\(\ \begin{array}{r} 8 x+2 y=12 \\ 5 x+2 y\ \ =9 \end{array}\)

Solve the system using elimination again. In this case, you can eliminate \(\ y\) by adding the opposite of the second equation:

\(\ \begin{array}{r} 8 x+\ \ \ 2 y &=&12 \\ -5 x+-2 y &=&-9 \\ \hline 3 x\ \ \ \ \ \ \ \ \ \ \ \ &=&3 \end{array}\)

Solve the resulting equation for the remaining variable.

\(\ \begin{array}{r} 3 x=3 \\ x=1 \end{array}\)

Now you use one of the equations in the two-variable system to find \(\ y\).

\(\ \begin{array}{r} 5 x+2 y=9 \\ 5(1)+2 y=9 \\ 5+2 y=9 \\ 2 y=4 \\ y=2 \end{array}\)

Finally, use any equation from the first system, along with the values already found, to solve for the last variable.

\(\ \begin{array}{r} 2 x-2 y+z=1 \\ 2(1)-2(2)+z=1 \\ 2-4+z=1 \\ -2+z=1 \\ z=3 \end{array}\)

Be sure to check your answer. With this many steps, there are a lot of places to make a simple error!

\(\ \begin{array}{r} 3 x+4 y-z=8 \\ 3(1)+4(2)-3=8 \\ 3+8-3=8 \\ 11-3=8 \\ 8=8 \\ \text { TRUE } \end{array}\)

\(\ \begin{array}{r} 5 x-2 y+z=4 \\ 5(1)-2(2)+3=4 \\ 5-4+3=4 \\ 1+3=4 \\ 4=4 \\ \text { TRUE } \end{array}\)

\(\ \begin{array}{r} 2 x-2 y+z=1 \\ 2(1)-2(2)+3=1 \\ 2-4+3=1 \\ -2+3=1 \\ 1=1 \\ \text { TRUE } \end{array}\)

Since \(\ x=1\), \(\ y=2\), and \(\ z=3\) is a solution for all three equations, it’s the solution for the system of equations. Just as two values can be written as an ordered pair, three values can be written as an ordered triplet: .

Solving a system of three variables

  • Choose two equations and use them to eliminate one variable.
  • Choose another pair of equations and use them to eliminate the same variable.
  • Use the resulting pair of equations from steps 1 and 2 to eliminate one of the two remaining variables.
  • Solve the final equation for the remaining variable.
  • Find the value of the second variable. Do this by using one of the resulting equations from steps 1 and 2 and the value of the found variable from step 4.
  • Find the value of the third variable. Do this by using one of the original equations and the values of the found variables from steps 4 and 5.
  • Check your answer in all three equations !

Solve for \(\ f\), \(\ g\), and \(\ h\).

\(\ \begin{array}{r} f+g+h&=&13 \\ f-h&=&-2 \\ -2 f+g&=&3 \end{array}\)

The solution is \(\ (f, g, h)=(2,7,4)\).

As with systems of two equations with two variables, you may need to add the opposite of one of the equations or even multiply one of the equations before adding in order to eliminate one of the variables.

Solve for \(\ x\), \(\ y\), and \(\ z\).

\(\ \begin{array}{r} 3 x-2 y+\ \ z=\ 12 \\ x+3 y+\ \ z=-4 \\ 2 x+2 y-4 z=\ \ \ 6 \end{array}\)

The solution is \(\ (x, y, z)=(3,-2,-1)\).

These systems can be helpful for solving real-world problems.

Andrea sells photographs at art fairs. She prices the photos according to size: small photos cost $10, medium photos cost $15, and large photos cost $40. She usually sells as many small photos as medium and large photos combined. She also sells twice as many medium photos as large. A booth at the art fair costs $300.

If her sales go as usual, how many of each size photo must she sell to pay for the booth?

If Andrea sells small photos, medium photos, and large photos, she’ll receive exactly the amount of money needed to pay for the booth.

Systems with No Solutions or an Infinite Number of Solutions

Now let’s look at Case 2 (no solution) and Case 3 (an infinite number of solutions).

Since you will not graph these equations, as it is difficult to graph in three dimensions on a 2-dimensional sheet of paper, you will look at what happens when you try to solve systems with no solutions or an infinite number of solutions.

Let’s look at a system that has no solutions.

\(\ \begin{array}{r} 5 x-2 y+\ \ z=\ \ \ 3 \\ 4 x-4 y-8 z=\ \ \ 2 \\ -x+\ \ y+2 z=-3 \end{array}\)

Suppose you wanted to solve this system, and you started with the last two equations. Multiply the third equation by 4 and add it to the second equation to eliminate \(\ x\).

\(\ \begin{array}{r} 4 x-4 y-8 z\ \ &=\ \ \ \ \ \ \ \ 2 \\ 4(-x+\ \ y+2 z) &=4(-3) \end{array}\)

\(\ \begin{array}{r} 4 x-4 y-8 z=\ \ \ \ \ 2 \\ -4 x+4 y+8 z=-12 \\ \hline 0=-10 \end{array}\)

In this case, the result is a false statement. This means there are no solutions to the two equations and therefore there can be no solutions for the system of three equations. If this occurs for any two of the three equations, then there is no solution for the system of equations.

Now let’s look at a system that has an infinite number of solutions.

\(\ \begin{array}{r} x-2 y+\ \ z=\ \ \ 3 \\ -3 x+6 y-3 z=-9 \\ 4 x-8 y+4 z=\ 12 \end{array}\)

For the first step, you would choose two equations and combine them to eliminate a variable. You can eliminate \(\ x\) by multiplying the first equation by 3 and adding to the second equation.

\(\ \begin{array}{r} 3(x-2 y+z)=\ 3(3) \\ -3 x+6 y-3 z=\ -9 \\ \hline \end{array}\)

\(\ \begin{array}{r} 3 x-6 y+3 z=\ \ \ 9 \\ -3 x+6 y-3 z= -9 \\ \hline 0=\ \ \ 0 \end{array}\)

Notice that when the two equations are added, all variables are eliminated! The final equation is a true statement: \(\ 0=0\).

When this happens, it’s because the two equations are equivalent . These two equations would graph as the same plane. But in order for the solution to the system of three equations to be infinite, you need to continue to check with the third equation.

Since the first two equations are equivalent, the system of equations could be written with only two equations. Continue as before. Multiply the first equation by -4 and add the third equation.

\(\ 4 x-8 y+4 z=-12\)

\(\ \begin{array}{r} -4 x+8 y-4 z= & -12 \\ 4 x-8 y+4 z= & 12 \\ \hline 0= & 0 \end{array}\)

Again, the final equation is the true statement \(\ 0=0\). So the third equation is the same plane as the first two. Now you can confirm that there are an infinite number of solutions—all of the points that are on the plane that these three equations each describe.

This is one type of situation where there are an infinite number of solutions. There are others, which you will not examine at this time.

How many solutions does the following system of equations have?

\(\ \begin{array}{c} x+y+z=2 \\ 2 x+2 y+2 z=4 \\ -3 x-3 y-3 z=-6 \end{array}\)

There are an infinite number of solutions to this system.

Solve the following system of equations.

\(\ \begin{array}{c} x-y-2 z=4 \\ 4 x-4 y-z=2 \\ -x+y+2 z=-3 \end{array}\)

The system has no solutions.

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Mathematics LibreTexts

3.04: Newton-Raphson Method for Solving a Nonlinear Equation

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Lesson 1: Tangent to a Curve

Learning objectives.

After successful completion of this lesson, you should be able to

1) find the first derivative of a function,

2) calculate the value of the first derivative of the function at a particular point,

3) relate the first derivative of the function at a specific point to the slope of the tangent line, and,

4) find where the tangent line crosses the \(x\) -axis.

Introduction

The formula for solving a nonlinear equation of the form \(f(x) = 0\) by the Newton-Raphson method is given by

\[x_{i + 1} = x_{i} - \frac{f(x_{i})}{f^{\prime}(x_{i})} \nonumber\]

Some of the pre-requisites to learning the Newton-Raphson method are

a) finding the first derivative of a function,

b) calculating the value of the first derivative of the function at a particular point,

c) relating the first derivative of the function at a specific point to the slope of the tangent line, and

d) finding where the tangent line crosses the x-axis.

Let us illustrate the above concepts through an example.

Example \(\PageIndex{1.1}\)

For \(f(x) = 0.13x^{3}\) , find the following

a) Find \(f^{\prime}(x)\) .

b) Find \(f^{\prime}(7)\) .

c) Find the slope of the tangent line to the function at \ (x = 7\) .

d) Find the angle the tangent line makes to the \(x\)-axis at \(x = 7\) . Write your answer both in degrees and radians.

e) Find where the tangent line to the function at \(x = 7\) crosses the \(x\)-axis.

a) \[f(x) = 0.13x^{3} \nonumber\]

\[\frac{d}{dx}(x^{n}) = nx^{n - 1}, n \neq 0 \nonumber\]

\[\frac{d}{dx}(aq(x)) = a\frac{dq}{dx} \nonumber\]

where \(a\) is a constant, we get

\[\begin{split} f^{\prime}(x) &= \frac{d}{dx}(0.13x^{3})\\ &= 0.13\frac{d}{dx}(x^{3})\\ &= 0.13(3)x^{2}\\ &= 0.39x^{2}\end{split} \nonumber\]

b) From part (a)

\[f^{\prime}(x) = 0.39x^{2} \nonumber\]

\[\begin{split} f^{\prime}(7) &= 0.39(7)^{2}\\ &= 19.11\end{split} \nonumber\]

c) From part (b)

\[f^{\prime}(7) = 19.11 \nonumber\]

The slope of the tangent line to the function at \(x = 7\) is \(f^{\prime}(7)\) and that is \(19.11\) .

d) From part (c), the slope of the tangent to the function at \(x = 7\) is \(19.11\) .

Graph of the function y=0.13 x^2, showing the tangent line to this function at x=7 and the angle theta that the tangent line makes when crossing the x-axis.

\[\tan\theta = 19.11 \nonumber\]

\[\theta = \text{angle made by the tangent line to the}\ x \text{-axis.} \nonumber\]

\[\begin{split} \theta &= \tan^{- 1}(19.11)\\ &= \ 87.004{^\circ} \text{ or } 1.5185 \text{ radians}\end{split} \nonumber\]

e) The equation of a straight line is

\[y = mx + c \nonumber\]

Since we know the slope of the tangent line, that is, \(19.11\) , and we know one of the points on the tangent line is \((7,f(7)) = (7,0.13 \times 7^{3}) = (7,44.59)\), we can find the intercept of the straight line

\[44.59 = 19.11(7) + c \nonumber\]

\[c = - 89.18 \nonumber\]

Hence the equation of the tangent line is

\[y = 19.11x - 89.18 \nonumber\]

The tangent line crosses the \(x\) - axis when \(y=0\) , that is,

\[0 = 19.11x - 89.18 \nonumber\]

\[\begin{split} x &= \frac{89.18}{19.11}\\ &= 4.667\end{split} \nonumber\]

Audiovisual Lectures

Title: Background of Newton-Raphson Method.

Summary : Learn the background Newton Raphson method of solving a nonlinear equation of the form \(f(x) = 0\) . This includes finding the slope of the tangent line to the function at a particular point, finding the angle the tangent line makes to the \(x\)-axis, and finding where the tangent line to the function crosses the \(x\)-axis.

Lesson 2: Graphical Derivation of Newton-Raphson Method

1) derive the Newton-Raphson method to solve a nonlinear equation

2) write the algorithm for the Newton-Raphson method to solve a nonlinear equation

Methods such as the bisection method and the false position method of finding roots of a nonlinear equation \(f(x) = 0\) require bracketing of the root by two guesses. Such methods are called bracketing methods . These methods are always convergent since they are based on reducing the interval between the two guesses so as to zero in on the root of the equation.

In the Newton-Raphson method, the root is not bracketed. In fact, only one initial guess of the root is needed to get the iterative process started to find the root of an equation. The method hence falls in the category of open methods . Convergence in open methods is not guaranteed, but it does so much faster than the bracketing methods if the method does converge.

The Newton-Raphson method is based on the principle that if the initial guess of the root of \(f(x) = 0\) is at \(x_{i}\) , then if one draws the tangent to the curve at \((x_i,f(x_{i})\) , the point \(x_{i + 1}\) where the tangent crosses the \(x\) -axis is an improved estimate of the root (Figure \(\PageIndex{2.1}\)). One can then use \(x_{i+1}\) as the next point to draw the tangent line to the function \(f(x)\) and find out where that tangent line crosses the \(x\) -axis. Continuing this process brings us closer and closer to the root of the equation.

Geometrical illustration of the Newton-Raphson method.

Using the definition of the slope of a function, at \(x = x_{i}\)

\[\begin{split} f^{\prime}\left( x_{i} \right)\ &= \ \tan\theta\\ &= \ \frac{f\left( x_{i} \right) - 0}{x_{i} - x_{i + 1}},\end{split} \nonumber\]

which gives

\[x_{i + 1}\ = \ x_{i} - \frac{f\left( x_{i} \right)}{f^{\prime}\left( x_{i} \right)}\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.1}) \nonumber\]

Equation \((\PageIndex{2.1})\) is called the Newton-Raphson formula for solving nonlinear equations of the form \(f\left( x \right) = 0\) . So starting with an initial guess, \(x_{i}\) , one can find the next guess, \(x_{i + 1}\), by using Equation \((\PageIndex{2.1})\). One can repeat this process until one finds the root within a desirable tolerance.

The steps of the Newton-Raphson method to find the root of an equation \(f\left( x \right) = 0\) are

  • Evaluate \(f^{\prime}\left( x \right)\) symbolically
  • Use an initial guess of the root, \(x_{i}\) , to estimate the new value of the root, \(x_{i + 1}\) , as

\[x_{i + 1}\ = \ x_{i} - \frac{f\left( x_{i} \right)}{f^{\prime}\left( x_{i} \right)} \nonumber\]

  • Find the absolute relative approximate error \(\left| \epsilon_{a} \right|\) as

\[\left| \epsilon_{a} \right|\ = \ \left| \frac{x_{i + 1} - \ x_{i}}{x_{i + 1}} \right| \times 100\;\;\;\;\;\;\;\;\;\;\;\; (\PageIndex{2.2}) \nonumber\]

  • Compare the absolute relative approximate error with the pre-specified relative error tolerance, \(\epsilon_{s}\) . If \(\left| \epsilon_{a} \right| > \epsilon_{s}\) , then go to Step 2, else stop the algorithm. Also, check if the number of iterations has exceeded the maximum number of iterations allowed. If so, one needs to terminate the algorithm and notify the user.

Title: Derivation of Newton-Raphson Method

Summary : Learn how to derive the Newton Raphson method of solving a nonlinear equation of the form \(f(x) = 0\) .

Lesson 3: Derivation of Newton-Raphson Method from Taylor Series

1) derive the Newton-Raphson method from the Taylor series

The Newton-Raphson method can also be derived from the Taylor series. For a general function \(f\left( x \right)\) , the Taylor series around the point \(x_{i}\) is

\[f\left( x_{i + 1} \right) = f\left( x_{i} \right) + f^{\prime}\left( x_{i} \right)\left( x_{i + 1} - x_{i} \right)+ \frac{f^{\prime\prime}\left( x_{i} \right)}{2!}\left( x_{i + 1} - x_{i} \right)^{2} + \ldots \nonumber\]

We are seeking a point where \(f\left( x \right) = 0\); that is, if we assume

\[f\left( x_{i + 1} \right) = 0, \nonumber\]

\[0 = f\left( x_{i} \right) + f^{\prime}\left( x_{i} \right)\left( x_{i + 1} - x_{i} \right) + \frac{f^{\prime\prime}\left( x_{i} \right)}{2!}\left( x_{i + 1} - x_{i} \right)^{2} + \ \ldots \nonumber\]

Using only the first two terms of the series gives

\[0 \approx f\left( x_{i} \right) + f^{\prime}\left( x_{i} \right)\left( x_{i + 1} - x_{i} \right) \nonumber\]

Solving for \(x_{i + 1}\) gives

\[x_{i + 1} \approx x_{i} - \frac{f\left( x_{i} \right)}{f^{\prime}\left( x_{i} \right)} \nonumber\]

This is the same Newton-Raphson method formula as derived in the previous lesson using geometry and differential calculus.

Title: Newton-Raphson Method Derivation from Taylor Series

Summary : This video discusses how to derive Newton Raphson method from Taylor's theorem.

Lesson 4: Application of Newton-Raphson Method

1) apply the Newton-Raphson method to solve for roots of a nonlinear equation.

Applications

In the previous lessons, we discussed the background to the Newton-Raphson Method of solving nonlinear equations. In this lesson, we take an example of how to apply the algorithm of the Newton-Raphson method to solve a nonlinear equation.

Example \(\PageIndex{4.1}\)

Solve the nonlinear equation \(x^{3} = 20\) by the Newton-Raphson method using an initial guess of \(x_{0} = 3\) .

Conduct three iterations to estimate the root of the above equation.

Find the absolute relative approximate error at the end of each iteration.

Find the number of significant digits at least correct at the end of each iteration.

First, we rewrite the equation \(x^{3} = 20\) in the form \(f(x) = 0\)

\[f(x) = x^{3} - 20 \nonumber\]

\[f^{\prime}(x) = 3x^{2} \nonumber\]

The initial guess is

\[x_{0} = 3 \nonumber\]

Iteration 1

The estimate of the root is

\[\begin{split} x_{1} &= x_{0} - \frac{f(x_{0})}{f^{\prime}(x_{0})}\\ &= 3 - \frac{3^{3} - 20}{(3)^{2}}\\ &= 3 - \frac{7}{27}\\ &= 3 - 0.259259\\ &= 2.74074\end{split} \nonumber\]

The absolute approximate relative error \(\left| \epsilon_{a} \right|\) at the end of Iteration 1 is

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{x_{1} - x_{o}}{x_{1}} \right| \times 100\\ &= \left| \frac{2.74074 - 3}{2.74074} \right| \times 100\\ &= 9.46\%\end{split} \nonumber\]

The number of significant digits at least correct is \(0\) , as you need an absolute relative approximate error of \(5 \%\) or less for at least one significant digit to be correct in your result.

Iteration 2

\[\begin{split} x_{2} &= x_{1} - \frac{f(x_{1})}{f^{\prime}(x_{1})}\\ &= 2.74074 - \frac{(2.74074)^{3} - 20}{3(2.74074)^{2}}\\ &= 2.74074 - \frac{0.587495}{22.5350}\\ &= 2.71467 \end{split} \nonumber\]

The absolute relative approximate error \(\left| \epsilon_{a} \right|\) at the end of Iteration 2 is

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{x_{2} - x_{1}}{x_{2}} \right| \times 100\\ &= \left| \frac{2.71467 - 2.74074}{2.71467} \right| \times 100\\ &= 0.960\% \end{split} \nonumber\]

The maximum value of \(m\) for which \(\left| \epsilon_{a} \right|\leq0.5 \times 10^{2 - m}\%\) is \(1.71669\) . Hence, the number of significant digits at least correct is \(1\) .

Iteration 3

\[\begin{split} x_{3} &= x_{2} - \frac{f(x_{2})}{f^{\prime}(x_{2})}\\ &= 2.71467 - \frac{(2.71467)^{3} - 20}{(2.71476)^{2}}\\ &= 2.71467 - \frac{5.57925 \times 10^{- 3}}{22.1083}\\ &= 2.71467 - 2.52360 \times 10^{- 4}\\ &= 2.71442\end{split} \nonumber\]

The absolute relative approximate error \(\left| \epsilon_{a} \right|\) at the end of Iteration 3 is

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{2.71442 - 2.71467}{2.71442} \right| \times 100\\ &= 0.00921\% \end{split} \nonumber\]

The maximum value of \(m\) for which \(\ \left| \epsilon_{a} \right|\leq 0.5 \times 10^{2 - m}\%\) is \(\ 3.73471\) . Hence, the number of significant digits at least correct is \(3\) .

Example \(\PageIndex{4.2}\)

You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The floating ball has a specific gravity of \(0.6\) and has a radius of \(5.5\ \text{cm}\) . You are asked to find the depth to which the ball is submerged when floating in the water.

A ball of radius R floats in water, with a region of the ball of height x submerged.

The equation that gives the depth \(x\) in meters to which the ball is submerged underwater is given by

\[x^{3} - 0.165x^{2} + 3.993 \times 10^{- 4} = 0 \nonumber\]

Use the Newton-Raphson method of finding roots of equations to find

a) the depth \(x\) to which the ball is submerged underwater. Conduct three iterations to estimate the root of the above equation.

b) the absolute relative approximate error at the end of each iteration, and

c) the number of significant digits at least correct at the end of each iteration.

\[f\left( x \right) = x^{3} - 0.165x^{2} + 3.993 \times 10^{- 4} \nonumber\]

\[f^{\prime}\left( x \right) = 3x^{2} - 0.33x \nonumber\]

Let us assume the initial guess of the root of \(f\left( x \right) = 0\) is \(x_{0} = 0.05\text{ m}.\) This is a reasonable guess (discuss why \(x = 0\) and \(x = 0.11\text{ m}\) are not good choices) as the extreme values of the depth \(x\) would be \(0\) and the diameter ( \(0.11\ \text{m}\) ) of the ball.

\[\begin{split} x_{1} &= x_{0} - \frac{f\left( x_{0} \right)}{f^{\prime}\left( x_{0} \right)}\\ &= 0.05 - \frac{\left( 0.05 \right)^{3} - 0.165\left( 0.05 \right)^{2} + 3.993 \times 10^{- 4}}{3\left( 0.05 \right)^{2} - 0.33\left( 0.05 \right)}\\ &= 0.05 - \frac{1.118 \times 10^{- 4}}{- 9 \times 10^{- 3}}\\ &= 0.05 - \left( - 0.01242 \right)\\ &= 0.06242 \end{split} \nonumber\]

The absolute relative approximate error \(\left| \epsilon_{a} \right|\) at the end of Iteration 1 is

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{x_{1} - x_{0}}{x_{1}} \right| \times 100\\ &= \left| \frac{0.06242 - 0.05}{0.06242} \right| \times 100\\ &=19.90\% \end{split} \nonumber\]

The number of significant digits at least correct is \(0\) , as you need an absolute relative approximate error of \(5\%\) or less for at least one significant digit to be correct in your result.

\[\begin{split} x_{2} &= x_{1} - \frac{f\left( x_{1} \right)}{f^{\prime}\left( x_{1} \right)}\\ &= 0.06242 - \frac{\left( 0.06242 \right)^{3} - 0.165\left( 0.06242 \right)^{2} + 3.993 \times 10^{- 4}}{3\left( 0.06242 \right)^{2} - 0.33\left( 0.06242 \right)}\\ &= 0.06242 - \frac{- 3.97781 \times 10^{- 7}}{- 8.90973 \times 10^{- 3}}\\ &= 0.06242 - \left( 4.4646 \times 10^{- 5} \right)\\ &= 0.06238 \end{split} \nonumber\]

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{x_{2} - x_{1}}{x_{2}} \right| \times 100\\ &= \left| \frac{0.06238 - 0.06242}{0.06238} \right| \times 100\\ &= 0.0716\% \end{split} \nonumber\]

The maximum value of \(m\) for which \(\left| \epsilon_{a} \right| \leq 0.5 \times 10^{2 - m}\) is \(2.844\) . Hence, the number of significant digits at least correct in the answer is \(2\) .

\[\begin{split} x_{3} &= x_{2} - \frac{f\left( x_{2} \right)}{f^{\prime}\left( x_{2} \right)}\\ &= 0.06238 - \frac{\left( 0.06238 \right)^{3} - 0.165\left( 0.06238 \right)^{2} + 3.993 \times 10^{- 4}}{3\left( 0.06238 \right)^{2} - 0.33\left( 0.06238 \right)}\\ &= 0.06238 - \frac{4.44 \times 10^{- 11}}{- 8.91171 \times 10^{- 3}}\\ &= 0.06238 - \left( - 4.9822 \times 10^{- 9} \right)\\ &= 0.06238 \end{split} \nonumber\]

\[\begin{split} \left| \epsilon_{a} \right| &= \left| \frac{0.06238 - 0.06238}{0.06238} \right| \times 100\\ &= 0 \end{split} \nonumber\]

The number of significant digits at least correct is \(4\) , as only \(4\) significant digits are carried through in all the calculations.

Lesson 5: Advantages and Pitfalls of the Newton-Raphson Method

1) enumerate the advantages of the Newton-Raphson method and the reasoning behind them.

2) list the drawbacks of the Newton-Raphson method and the reason behind them.

Advantages of the Newton-Raphson Method

1) Fast convergence: Newton-Raphson method converges fast when it converges. The method converges quadratically – one can interpret that the number of significant digits correct gets approximately squared at each iteration.

2) Only one initial estimate is needed: Newton-Raphson method is an open method that requires only one initial estimate of the root of the equation. However, since the method is an open method, convergence is not guaranteed.

Drawbacks of the Newton-Raphson Method

1) Divergence at inflection point s: If the selection of the initial guess or an iterated value of the root turns out to be close to the inflection point of the function \(f\left( x \right)\) in the equation \(f\left( x \right) = 0\) , Newton-Raphson method may start diverging away from the root. It may then begin to converging back to the root. For example, to find the root of the equation

\[f\left( x \right) = \left( x - 1 \right)^{3} + 0.512 = 0 \nonumber\]

\[f^\prime(x) = 3(x-1)^2 \nonumber\]

and the Newton-Raphson method reduces to

\[x_{i + 1}\ = \ x_{i} - \frac{({x_{i}}^{3} - 1)^{3} + 0.512}{3(x_{i} - 1)^{2}} \nonumber\]

Starting with an initial guess of \(x_{0} = 5.0\) , Table \(\PageIndex{5.1}\) shows the iterated values of the root of the equation. As you can observe, the root starts to diverge at Iteration 6 because the previous estimate of \(0.92589\) is close to the inflection point of \(x = 1\) (the value of \(f^\prime\left( x \right)\) is zero at the inflection point). Eventually, after \(12\) more iterations, the root converges to the exact value of \(x = 0.2\) .

Divergence at an inflection point for f(x) = (x−1)^3 + 0.512 = 0.

2) Division by zer o: Division by zero may occur during the implementation of the Newton-Raphson method. For example, for the equation

\[f(x) = x^{3} - 0.03 x^{2} + 2.4 \times 10^{-6} = 0 \nonumber\]

\[f^\prime(x) = 3x^2-0.06x \nonumber\]

\[x_{i + 1}\ = \ x_{i} - \frac{x_{i}^{3} - 0.03x_{i}^{2} + 2.4 \times 10^{- 6}}{3x_{i}^{2} - 0.06x_{i}} \nonumber\]

For \(x_{0} = 0\) or \(x_{0} = 0.02\) , division by zero occurs (Figure \(\PageIndex{5.2}\)). For an initial guess close to \(0.02\) , such as \(x_{0} = 0.01999\) , one may avoid division by zero, but then the denominator in the formula is a small number. For this case of initial guess of \(x_{0}=0.01999\) , as given in Table \(\PageIndex{5.2}\), even after 9 iterations, the Newton-Raphson method does not converge.

Pitfall of division by zero or a near-zero number.

3) Oscillations near local maximum and minimum : Results obtained from the Newton-Raphson method may oscillate about the local maximum or minimum without converging on a root but converging on the local maximum or minimum. Eventually, it may lead to division by a number close to zero and may diverge.

For example, for

\[f\left( x \right) = x^{2} + 2 = 0 \nonumber\]

the equation has no real roots (Figure \(\PageIndex{5.3}\) and Table \(\PageIndex{5.3}\)).

Oscillations around local minima for f(x) = x^2 + 2.

4) Root jumping : In some cases where the function \(f(x)\) is oscillating and has several roots, one may choose an initial guess close to a root. However, the guesses may jump and converge to some other root. For example, for solving the equation \(\sin x = 0\) if you choose \(x_{0} = 2.4\pi = \left( 7.539822 \right)\) as an initial guess, it converges to the root of \(x = 0,\) as shown in Table 4 and Figure 4. However, one may have chosen this as an initial guess to converge to \(x = 2\pi = 6.2831853\) .

Root jumping from intended location of root for f(x) = sin(x)= 0.

Title: Newton-Raphson Method: Advantages and Drawbacks

Summary : This video discusses the advantages and drawbacks of Newton Raphson method of solving nonlinear equations.

Appendix A. What is an inflection point?

For a function \(f\left( x \right)\) , the point where the concavity changes from up-to-down or down-to-up is called its inflection point; that is, the sign of curvature changes sign. For example, for the function \(f\left( x \right) = \left( x - 1 \right)^{3}\) , the concavity changes at \(x = 1\) , and hence (1,0) is an inflection point.

For a function \(f\left( x \right)\) , for \((x,f\left( x \right))\) to be an inflection point, \(f^{\prime\prime}(x) = 0\) is a necessary condition. However, an inflection point may or may not exist at a point where \(f^{\prime\prime}(x) = 0\) . For example, for \(f(x) = x^{4} - 16\) , \(f^{\prime\prime}(0) = 0\) , but the concavity does not change at \(x = 0\) . Hence the point \((0, -16)\) is not an inflection point of \(f(x) = x^{4} - 16\) .

For \(f\left( x \right) = \left( x - 1 \right)^{3}\) , \(f^{\prime\prime}(1) = 0\) and \(f^{\prime\prime}(x)\) changes sign at \(x = 1\) ( \(f^{\prime\prime}(x) < 0\) for \(x = 1^{-}\) , and \(f^{\prime\prime}(x) > 0\) for \(x = 1^{+}\) ), and thus brings up the Inflection Point Theorem for a function \(f(x)\), which states the following.

“If \(f^\prime(c)\) exists and \(f^{\prime\prime}(c)\) changes sign at \(x = c\) , then the point \((c,f(c))\) is an inflection point of the graph of \(f\) .”

Multiple Choice Test

(1). The Newton-Raphson method of finding roots of nonlinear equations falls under the category of _____________ methods.

(A) bracketing

(D) graphical

(2). The Newton-Raphson method formula for finding the square root of a real number \(R\) from the equation \(x^{2} - R = 0\) is,

(A) \(\displaystyle x_{i + 1} = \frac{x_{i}}{2}\)

(B) \(\displaystyle x_{i + 1} = \frac{3x_{i}}{2}\)

(C) \(\displaystyle x_{i + 1} = \frac{1}{2}\left( x_{i} + \frac{R}{x_{i}} \right)\)

(D) \(\displaystyle x_{i + 1} = \frac{1}{2}\left( 3x_{i} - \frac{R}{x_{i}} \right)\)

(3). The next iterative value of the root of \(x^{2}-4=0\) using the Newton-Raphson method, if the initial guess is \(3\) , is

(A) \(1.5\)

(B) \(2.067\)

(C) \(2.167\)

(D) \(3.000\)

(4). The root of the equation \(f(x) = 0\) is found by using the Newton-Raphson method. The initial estimate of the root is \(x_{0} = 3\) , \(f\left( 3 \right) = 5\) . The angle made by the line tangent to the function \(f(x)\) makes at \(x = 3\) is \(57{^\circ}\) with respect to the \(x\) -axis. The next estimate of the root, \(x_{1}\) most nearly is

(A) \(-3.2470\)

(B) \(-0.24704\)

(C) \(3.2470\)

(D) \(6.2470\)

(5). The root of \(x^{3} = 4\) is found by using the Newton-Raphson method. The successive iterative values of the root are given in the table below.

The iteration number at which I would first trust at least two significant digits in the answer is

(6). The ideal gas law is given by

\(pv = RT\)

where \(p\) is the pressure, \(v\) is the specific volume, \(R\) is the universal gas constant, and \(T\) is the absolute temperature. This equation is only accurate for a limited range of pressure and temperature. Vander Waals came up with an equation that was accurate for a larger range of pressure and temperature, given by

\[ \left( p + \frac{a}{v^{2}} \right)\left( v - b \right) = RT \nonumber\]

where a and b are empirical constants dependent on a particular gas. Given the value of \(R = 0.08\) , \(a = 3.592\) , \(b = 0.04267\) , \(p = 10\) and \(T = 300\) (assume all units are consistent), one is going to find the specific volume, \(v\) , for the above values. Without finding the solution from the Vander Waals equation, what would be a good initial guess for \(v\) ?

(B) \(1.2\)

(C) \(2.4\)

(D) \(3.6\)

For complete solution, go to

http://nm.mathforcollege.com/mcquizzes/03nle/quiz_03nle_newton_solution.pdf

Problem Set

(1). Find the estimate of the root of \(x^{2} - 4 = 0\) by using the Newton-Raphson method. Assume that the initial guess of the root is \(3\) . Conduct three iterations. Also, calculate the approximate error, true error, absolute relative approximate error, and absolute relative true error at the end of each iteration.

(2). The velocity of a body is given by \(v(t) = 5e^{- t} + 6\) , where \(v\) is in m/s and \(t\) is in seconds.

a) Use the Newton Raphson method to find the time when the velocity will be \(7.0\) m/s. Use two iterations and take \(t = 2\) seconds as the initial guess.

b) What is the relative true error at the end of the second iteration for part (a)?

\(a)\ 1.5236,\ 1.6058\ \ b) \ 0.2260\%\)

(3). You are working for DOWN THE TOILET COMPANY that makes floats for ABC commodes. The ball has a specific gravity of \(0.6\) and has a radius of \(5.5\) cm. You are asked to find the depth to which the ball will get submerged when floating in the water.

The equation that gives the depth \(x\) (unit of \(x\) is meters) to which the ball is submerged under water is given by

Solving it exactly would require some effort. However, using numerical techniques such as Newton-Raphson method, we can solve this equation and any other equation of the form \(f(x) = 0\) . Solve the above equation by the Newton-Raphson method and do the following.

a) Conduct three iterations. Use an initial guess \(x = 0.055\ \text{m}\) .

b) Calculate the absolute relative approximate error at the end of each step.

c) Find the number of significant digits at least correct at the end of each iteration.

d) How could you have used the knowledge of the physics of the problem to develop initial guess for the Newton-Raphson method?

\(a)\ 0.06233,\ 0.06237,\ 0.06237\)

\(b)\ n/a,\ 0.06413\%,\ 0\%\) ( \(0\%\) if only 4 significant digits were carried in the calculations)

\(c)\ 0,\ 2,\ 4\) (as only 4 significant digits were carried in the calculations)

\(d)\) Since the diameter is \(0.11\ \text{m}\) , a value of \(0\) or \(0.11\) seems to be a good initial estimate, but these guesses give division by zero. So, \(0.055\) would be a good initial estimate.

(4). The root of the equation \(f(x) = 0\) is found by using the Newton-Raphson method. If the initial estimate of the root is assumed to be \(x_{0} = 3\) , given \(f(3) = 5\) , and the angle the tangent makes to the function \(f(x)\) at \(x = 3\) is \(57^{\circ}\) , what is the next estimate of the root, \(x_{1}\) ?

\(-0.24696\)

(5). The root of the equation \(f(x) = 0\) is found by using the Newton-Raphson method. The initial estimate of the root is assumed to be \(x_{0} = 5.0\) , and the angle the tangent makes with the \(x\) -axis to the function \(f(x)\) at \(x_{0} = 5.0\) is \(89.236^{\circ}\) . If the next estimate of the root is, \(x_{1} = 3.693\) , what is the value of the function \(f(x)\) at \(x = 5\) ?

(6). Find the Newton-Raphson method formula for finding the square root of a real number \(R\) from the equation \(x^{2} - R = 0\) . Find the square root of \(50.41\) by using the formula. Assume \(50.41\) as the initial guess. How many iterations does it take to get at least \(3\) significant digits correct in your answer? Do not use the true value to answer the question.

Takes 6 iterations. Root at the end of 6 th iteration is \(7.100\) , with the absolute relative approx error of \(0.014\%\) . Four significant digits were used in solving the problem.

IMAGES

  1. How to solve systems of linear equations on the SAT & ACT Math sections

    using systems of two equations to solve application problems

  2. 4 Ways to Solve Systems of Equations

    using systems of two equations to solve application problems

  3. How to Solve Systems of Equations by Graphing

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  4. Writing & Solving a System of Two Linear Equations Given a Table of

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  5. Solving Systems of Equations w/ 2-variables

    using systems of two equations to solve application problems

  6. Algebra 4-2: Solving Systems of Equations by Substitution

    using systems of two equations to solve application problems

VIDEO

  1. Lesson 6.3 Solve Systems of Equations by Substitution Part 4

  2. Systems of Equations Solve by Substitution

  3. Lesson 6.3 Solve Systems of Equations by Substitution Part 3

  4. Solving Application Problems Using Systems

  5. Solve A System Of Equations Using Elimination

  6. Solve Systems of 3-variable Equations

COMMENTS

  1. 4.2 Solve Applications with Systems of Equations

    Step 3. Name what we are looking for. Choose variables to represent those quantities. Step 4. Translate into a system of equations. Step 5. Solve the system of equations using good algebra techniques. Step 6. Check the answer in the problem and make sure it makes sense.

  2. 4.3: Solve Applications with Systems of Equations

    Step 5. Solve the system of equations. We will use substitution since the first equation is solved for a. Substitute 3b+103b+10 for a in the second equation. Solve for b. Substitute b=20b=20 into the first equation and then solve for a. Step 6. Check the answer in the problem. We will leave this to you! Step 7. Answer the question.

  3. 5.4 Solve Applications with Systems of Equations

    Choose variables to represent those quantities. Step 4. Translate into a system of equations. Step 5. Solve the system of equations using good algebra techniques. Step 6. Check the answer in the problem and make sure it makes sense. Step 7. Answer the question with a complete sentence.

  4. 4.4. Solve Applications with Systems of Equations

    Solve the system of equations using good algebra techniques. Check the answer in the problem and make sure it makes sense. Answer the question with a complete sentence. Translate to a System of Equations. Many of the problems we solved in earlier applications related two quantities.

  5. PDF Applications of Systems of Equations

    Steps for solving applications: The equations will usually be linear, but not always. Example 1: Set-up a system of equations and solve using any method. 25 miles 40 miles of gasoline in the city and on the highway. A recent sales gallon gallon trip that covered 1800 miles required 51 gallons of gasoline.

  6. 4.3 Solve Mixture Applications with Systems of Equations

    Step 5. Solve the system of equations We will use substitution to solve. Solve the first equation for b. Substitute b = −f + 21.540 into the second equation. Simplify and solve for f. To find b, substitute f = 12,870 into the first equation. Step 6. Check the answer in the problem. We leave the check to you. Step 7. Answer the question.

  7. 2.2: Solve Applications with Systems of Equations

    Solve applications with systems of equations. Read the problem. Make sure all the words and ideas are understood. Identify what we are looking for. Name what we are looking for. Choose variables to represent those quantities. Translate into a system of equations. Solve the system of equations using good algebra techniques.

  8. Solve Applications with Systems of Equations

    Solve Direct Translation Applications. Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we'll first translate the words into a system of linear equations.

  9. Solve Applications with Systems of Equations

    HOW TO: Solve applications with systems of equations. Read the problem. Make sure all the words and ideas are understood. Identify what we are looking for. Name what we are looking for. Choose variables to represent those quantities. Translate into a system of equations. Solve the system of equations using good algebra techniques.

  10. 4.4: Applications of 2×2 Systems of Equations

    Write a system of linear equations representing a distance, rate, and time problem. Distance, rate, and time applications are common in algebra. To solve this type of problem, we will use the following formula: distance = rate×time distance = rate × time. d =r⋅t d = r ⋅ t.

  11. 5.4: Solve Applications with Systems of Equations

    Solve the system of equations using good algebra techniques. ... Many of the problems we solved in earlier applications related two quantities. Here are two of the examples from the chapter on Math Models. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

  12. Applications of Systems

    Write a system of linear equations representing a mixture problem, solve the system and interpret the results. One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution.

  13. 4.4: Solve Mixture Applications with Systems of Equations

    Step 5. Solve the system of equations We will use elimination to solve the system. Multiply the first equation by \(−2\) to eliminate c. Simplify and add. Solve for n. To find the number of pounds of chocolate chips, substitute \(n=16\) into the first equation, then solve for c. Step 6. Check the answer in the problem.

  14. Systems of equations

    Systems of equations: trolls, tolls (2 of 2) Testing a solution to a system of equations. Systems of equations with graphing: y=7/5x-5 & y=3/5x-1. Systems of equations with graphing: exact & approximate solutions. Setting up a system of equations from context example (pet weights) Setting up a system of linear equations example (weight and price)

  15. 4.1 Solve Systems of Linear Equations with Two Variables

    We will solve larger systems of equations later in this chapter. An example of a system of two linear equations is shown below. We use a brace to show the two equations are grouped together to form a system of equations. { 2 x + y = 7 x − 2 y = 6. A linear equation in two variables, such as 2 x + y = 7, has an infinite number of solutions.

  16. Algebra

    Section 7.1 : Linear Systems with Two Variables. For problems 1 - 3 use the Method of Substitution to find the solution to the given system or to determine if the system is inconsistent or dependent. x−7y = −11 5x+2y = −18 x − 7 y = − 11 5 x + 2 y = − 18 Solution.

  17. 5.2

    Learning Objectives. (5.2.1) - Solve cost and revenue problems. Specify what the variables in a cost/ revenue system of linear equations represent. Determine and apply an appropriate method for solving the system. (5.2.2) - Solve value problems with a system of linear equations. (5.2.3) - Solve mixture problems with a system of linear ...

  18. Applications of Systems of Equations (examples, solutions, videos

    It is possible to solve word problems when two people are doing a work job together by solving systems of equations. To solve a work word problem, multiply the hourly rate of the two people working together times the time spent working to get the total amount of time spent on the job. Knowledge of solving systems of equations is necessary to ...

  19. 3.3: Applications of Linear Systems with Two Variables

    Use two variables as a means to simplify the algebraic setup of applications where the relationship between unknowns is unclear. Carefully read the problem several times. If two variables are used, then remember that you need to set up two linear equations in order to solve the problem.

  20. 2.9: Solving a System of Linear Equations

    To solve the system, though, you need two equations using two variables. Adding the first and third equations in the original system will also give an equation with x and y but not z. 3x + 4y − z = 8 2x − 2y + z = 1 5x + 2y = 9. Now you have a system of two equations and two variables. 8x + 2y = 12 5x + 2y = 9.

  21. Application of Systems of Equations Flashcards

    y = 2x + 1 and y + 6 = 3x. The sum of the digits of a two digit number is 14. The difference between the tens digit and the units digit is 2. If x is the tens digit and y is the ones digit, which system of equations represents the word problem? x + y = 14 and x - y = 2. A man bought 42 stamps, some 13¢ and some 18¢.

  22. 3.4: Applications of Systems

    One application of systems of equations are mixture problems. Mixture problems are ones where two different solutions are mixed together resulting in a new final solution. ... Let's use the problem solving process outlined in Module 1 to help us work through a solution to the problem. Read and Understand: We are looking for a new amount ...

  23. 4.4: Applications of Linear Systems

    We obtain the following equivalent system: Add the equations and then solve for x: Back substitute. x + y = 8 4.4+ y = 8 4.4 + y− 4.4 = 8− 4.4 x = 3.6. Answer: To obtain 8 ounces of a 32 % alcohol mixture we need to mix 4.4 ounces of the 50 % alcohol solution and 3.6 ounces of the 10 % solution.

  24. 3.04: Newton-Raphson Method for Solving a Nonlinear Equation

    Introduction. Methods such as the bisection method and the false position method of finding roots of a nonlinear equation \(f(x) = 0\) require bracketing of the root by two guesses. Such methods are called bracketing methods.These methods are always convergent since they are based on reducing the interval between the two guesses so as to zero in on the root of the equation.