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

[latex]S[/latex] = number of small photos sold

[latex]M[/latex] = number of medium photos sold

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

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

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

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

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

[latex]S=M+L[/latex]

She sells twice as many medium photos as large photos.

[latex]M=2L[/latex]

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 [latex]L[/latex] by adding [latex]M-2L=0[/latex] 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 [latex]M=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 equations 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.