Systems of Linear Equations: Three Variables: Apply It 1

Giselle Miller

Systems of Linear Equations: Three Variables: Apply It 1

Solve three equations with three different variables
Figure out when a set of three equations has no solution
Find and explain when a set of three equations has infinitely many solutions

Applications of Systems of Three Equations in Three Variables

Now we are ready to handle the problem we encountered as we began this section by using what we know about linear equations to translate the situation into a system of three equations. Then, we’ll use our new understanding of three-by-three systems to find the solution.

John received an inheritance of [latex]$12,000[/latex] that he divided into three parts and invested in three ways: in a money-market fund paying [latex]3\%[/latex] annual interest; in municipal bonds paying [latex]4\%[/latex] annual interest; and in mutual funds paying [latex]7\%[/latex] annual interest. John invested [latex]$4,000[/latex] more in municipal funds than in municipal bonds. He earned [latex]$670[/latex] in interest the first year. How much did John invest in each type of fund?

To solve this problem, we use all of the information given and set up three equations. First, we assign a variable to each of the three investment amounts:

[latex]\begin{align}&x=\text{amount invested in money-market fund} \\ &y=\text{amount invested in municipal bonds} \\ z&=\text{amount invested in mutual funds} \end{align}[/latex]

The first equation indicates that the sum of the three principal amounts is [latex]$12,000[/latex].

[latex]x+y+z=12{,}000[/latex]

We form the second equation according to the information that John invested [latex]$4,000[/latex] more in mutual funds than he invested in municipal bonds.

[latex]z=y+4{,}000[/latex]

The third equation shows that the total amount of interest earned from each fund equals [latex]$670[/latex].

[latex]0.03x+0.04y+0.07z=670[/latex]

Then, we write the three equations as a system.

[latex]\begin{align}x+y+z=12{,}000 \\ -y+z=4{,}000 \\ 0.03x+0.04y+0.07z=670 \end{align}[/latex]

To make the calculations simpler, we can multiply the third equation by [latex]100[/latex]. Thus,

[latex]\begin{align}x+y+z=12{,}000 \hspace{5mm} \left(1\right) \\ -y+z=4{,}000 \hspace{5mm} \left(2\right) \\ 3x+4y+7z=67{,}000 \hspace{5mm} \left(3\right) \end{align}[/latex]

Step 1. Interchange equation (2) and equation (3) so that the two equations with three variables will line up.

[latex]\begin{align}x+y+z=12{,}000\hfill \\ 3x+4y +7z=67{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

Step 2. Multiply equation (1) by [latex]-3[/latex] and add to equation (2). Write the result as row 2.

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ -y+z=4{,}000 \end{align}[/latex]

Step 3. Add equation (2) to equation (3) and write the result as equation (3).

[latex]\begin{align}x+y+z=12{,}000 \\ y+4z=31{,}000 \\ 5z=35{,}000 \end{align}[/latex]

Step 4. Solve for [latex]z[/latex] in equation (3). Back-substitute that value in equation (2) and solve for [latex]y[/latex]. Then, back-substitute the values for [latex]z[/latex] and [latex]y[/latex] into equation (1) and solve for [latex]x[/latex].

[latex]\begin{align}&5z=35{,}000 \\ &z=7{,}000 \\ \\ &y+4\left(7{,}000\right)=31{,}000 \\ &y=3{,}000 \\ \\ &x+3{,}000+7{,}000=12{,}000 \\ &x=2{,}000 \end{align}[/latex]

John invested [latex]$2,000[/latex] in a money-market fund, [latex]$3,000[/latex] in municipal bonds, and [latex]$7,000[/latex] in mutual funds.

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?

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]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.

[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 [latex]S[/latex] variable, it may be helpful to use the other two equations and eliminate the [latex]S[/latex] 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 [latex]S[/latex] 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 [latex]L[/latex], 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 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.