solving problems using linear systems

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?

(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.

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.

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.

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.

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].

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].

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.

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.

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.

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Systems of Linear Equations

Solve Systems of Linear Equations with Two Variables

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

Before you get started, take this readiness quiz.

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 .

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.

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.

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.

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

• Graph the first equation.
• Graph the second equation on the same rectangular coordinate system.
• 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.
• 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.

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.

no solution

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.

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

infinitely many solutions

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 have the same slope and same y- intercept.

The systems of equations in (Figure) and (Figure) each had two intersecting lines. Each system had one solution.

In (Figure) , 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 (Figure) , has no solution. We call a system of equations like this inconsistent. It has no solution.

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 (Figure) .

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

ⓐ 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.

ⓐ no solution, inconsistent, independent ⓑ one solution, consistent, independent

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.

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

Be very careful with the signs in the next example.

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

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.

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.

The steps are listed here for easy reference.

• 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.
• Add the equations resulting from Step 2 to eliminate one variable.
• Solve for the remaining variable.
• Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.

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.

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.

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.

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.

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.

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.

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.

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

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.

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

ⓐ Since both equations are in standard form, using elimination will be most convenient. ⓑ Since one equation is already solved for x , using substitution will be most convenient.

ⓐ Since one equation is already solved for y , using substitution will be most convenient. ⓑ Since both equations are in standard form, using elimination will be most convenient.

Key Concepts

Practice Makes Perfect

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

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

infinite solutions

1 point, consistent and independent

infinite solutions, consistent, dependent

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

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

infinitely many

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

ⓐ substitution ⓑ elimination

ⓐ elimination ⓑ substituion

Writing Exercises

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

Answers will vary.

ⓐ by graphing ⓑ by substitution

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. Who 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.

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

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Writing Systems of Linear Equations from Word Problems

As we learn about math, we often ask ourselves, "How is any of this knowledge useful in life? When will I ever need to use algebra?" Here's the thing: When you know how to "translate" word problems into algebraic equations, you'll immediately see how useful math really is. With a few simple steps, you can turn real questions about the world around you into equations, allowing you to calculate things that you never thought possible. It's one of the most interesting things about math -- so let's get started!

How to tell when a word problem can become a linear equation

First, we need to keep our eyes open for a number of clues. These clues tell us that we can turn our word problem into a linear equation:

• There are different quantities of things, such as a specific number of people, objects, hours, and so on.
• Each quantity has a clear value. Instead of saying, "There are a few boxes," the word problem needs to tell us how many boxes there are. Are there five? Six? Seven?
• We need to know at least some of these values to build our linear equation. The unknowns can become variables, like "x" or "y."

How to turn word problems into linear equations

If we follow a few simple steps, we can turn certain word problems into linear equations:

• Take a second to think about the "problem." What are we trying to find out? What is the "variable" in this word problem? Define all of the words carefully. If we can't define the words properly, our equation won't be accurate.
• Turn the word problem into an equation. Plug in all of your known values and use letters like "x" and "y" to represent the unknown variables. Make sure you write out what each variable represents below the equation so you don't forget.
• Solve the equation. Using our math skills, we can now solve the problem and find the values of our variables. We can use a wide range of strategies to solve the equation, including substitution, elimination, or graphing.

An example of a word problem translated into a linear equation

Now let's see how this all works with an example. Here's our word problem:

We decided to go to a music concert with all our friends, including 12 children and 3 adults. We paid for everyone's tickets for a total of $162. Another group of friends paid $122 for 8 children and 3 adults. How much does a child's ticket cost, and how much does an adult's ticket cost?

1. Understand the problem

We know two values: 12 children and 3 adults cost $162, while 8 children and 3 adults cost $122.

What we don't know is how much an adult's ticket costs, and how much a child's ticket costs.

Let's create variables for those unknowns:

x = the cost of one child's ticket

y = the cost of one adult's ticket

2. Translate the problem into an equation

We know that 12 children and 3 adults cost $162. Let's plug in our variables and create an equation based on this:

12x + 3y = 162

We can do the same for the other group of concert-goers:

8x + 3y = 122

3. Solve the equation

On one weekend they sold a total of 12 adult tickets and 3 child tickets for a total of 162 dollars, and the next weekend they sold 8 adult tickets and 3 child tickets for 122 dollars, find the price for a child's and adult's ticket.

When we have two very similar equations like this, we can simply subtract them from each other to get the values we need:

Now that we know the value of x, we can use it to find y.

12(10) + 3y = 162

120 + 3y = 162

Now we know that a child's ticket costs $10, while an adult's ticket costs $14.

We can now check our work by plugging our solutions back into our original equations:

12(10) + 3(14) = 162

8(10) + 3(14) = 122

You can use this strategy to solve all kinds of everyday math problems you encounter in life!

Topics related to the Writing Systems of Linear Equations from Word Problems

Consistent and Dependent Systems

Word Problems

Cramer's Rule

Flashcards covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Flashcards

College Algebra Flashcards

Practice tests covering the Writing Systems of Linear Equations from Word Problems

Algebra 1 Diagnostic Tests

College Algebra Diagnostic Tests

Get more help with linear equation word problems

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SAT (Fall 2023)

Course: sat (fall 2023)   >   unit 10.

• Solving linear equations and linear inequalities — Basic example
• Solving linear equations and linear inequalities — Harder example
• Interpreting linear functions — Basic example
• Interpreting linear functions — Harder example
• Linear equation word problems — Basic example
• Linear equation word problems — Harder example
• Linear inequality word problems — Basic example
• Linear inequality word problems — Harder example
• Graphing linear equations — Basic example
• Graphing linear equations — Harder example
• Linear function word problems — Basic example
• Linear function word problems — Harder example
• Systems of linear inequalities word problems — Basic example
• Systems of linear inequalities word problems — Harder example

Solving systems of linear equations — Basic example

• Solving systems of linear equations — Harder example
• Systems of linear equations word problems — Basic example
• Systems of linear equations word problems — Harder example

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