Applications of Linear Equations in Three Variables

Learning Outcomes

Solve applications using systems of equations in three variables

Systems can be helpful for solving real-world problems. In this example, we will write three equations that model sales at an art fair to learn how many prints should be sold to break even for the cost of the booth rental.

Example

Andrea sells photographs at art fairs. She prices the photos according to size: small photos cost [latex]$10[/latex], medium photos cost [latex]$15[/latex], and large photos cost [latex]$40[/latex]. 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 [latex]$300[/latex].

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

Show Solution

To set up the system, first choose the variables. In this case the unknown values are the number of small, medium, and large photos.

S = number of small photos sold

M = number of medium photos sold

L = number of large photos sold

The total of her sales must be [latex]$300[/latex] to pay for the booth. We can find the total by multiplying the cost for each size by the number of that size sold.

[latex]10[/latex]S = money received for small photos

[latex]15[/latex]M = money received for medium photos

[latex]40[/latex]L = money received for large photos

Total Sales:[latex]10[/latex]S +[latex]15[/latex]M +[latex]40[/latex]L =[latex]300[/latex]

The number of small photos is the same as the total of medium and large photos combined.

S = M + L

She sells twice as many medium photos as large photos.

M =[latex]2[/latex]L

To make things easier, rewrite the equations to be in the same format, with all variables on the left side of the equal sign and only a constant number on the right.

[latex]\begin{cases}10S+15M+40L=300\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,S–M–L=0\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,M–2L=0\end{cases}[/latex]

Now solve the system.

Step 1: First choose two equations and eliminate a variable. Since one equation has no S variable, it may be helpful to use the other two equations and eliminate the S variable from them. Multiply both sides of the second equation by [latex]−10[/latex].

[latex]\begin{array}{l}-10(S–M–L)=-10(0)\\-10s+10M+10L=0\end{array}[/latex]

Now add this modified equation with the first equation in the original list of equation.

[latex]\begin{array}{ccc}10S+15M+40L=300\\\underline{+(-10s+10M+10L=0)}\\25M+50L=300\end{array}[/latex]

Step 2: The other equation for our two-variable system will be the remaining equation (that has no S variable). Eliminate a second variable using the equation from step [latex]1[/latex]. While you could multiply the third of the original equations by [latex]25[/latex] to eliminate L, the numbers will stay nicer if you divide the resulting equation from step [latex]1[/latex] by [latex]25[/latex]. Do not forget to be careful of the signs!

Divide first:

[latex]\begin{array}{ccc}\dfrac{25}{25}M+\dfrac{50}{25}L=\dfrac{300}{25}\\M+2L=12\end{array}[/latex]

Now eliminate L by adding M-2L=0 to this new equation.

[latex]\begin{array}{l}M+2L=12\\\underline{M–2L=0}\\2M=12\\M=\dfrac{12}{2}=6\end{array}[/latex]

Step 3: Use M=[latex]6[/latex] and one of the equations containing just two variables to solve for the second variable. It is best to use one of the original equation in case an error was made in multiplication.

[latex]\begin{array}{ccc}M-2L=0\\6-2L=0\\-2L=-6\\L=3\end{array}[/latex]

Step 4: Use the two found values and one of the original equations to solve for the third variable.

[latex]\begin{array}{ccc}S–M–L=0\\S-6-3=0\\S-9=0\\S=9\end{array}[/latex]

Step 5: Check your answer. With application problems, it is sometimes easier (and better) to use the original wording of the problem rather than the equations you write.

She usually sells as many small photos as medium and large photos combined.

Medium and large photos combined: [latex]6 + 3 = 9[/latex], which is the number of small photos.

She also sells twice as many medium photos as large.

Medium photos is [latex]6[/latex], which is twice the number of large photos [latex](3)[/latex].

A booth at the art fair costs [latex]$300[/latex].

Andrea receives [latex]$10(9)[/latex] or [latex]$90[/latex] for the [latex]9[/latex] small photos, [latex]$15(6)[/latex] or [latex]$90[/latex] for the [latex]6[/latex] medium photos, and [latex]$40(3)[/latex] or [latex]$120[/latex] for the large photos. [latex]$90 + $90 + $120 = $300[/latex].

If Andrea sells [latex]9[/latex] small photos, [latex]6[/latex] medium photos, and [latex]3[/latex] large photos, she will receive exactly the amount of money needed to pay for the booth.

In the following video example, we show how to define a system of three equations in three variables that represents a mixture needed by a chemist.

Our last example shows you how to write a system of three equations that represents ticket sales for a theater that has three different prices for tickets.

Module 6: Linear Systems