4.4 Solve Applications with Systems of Equations

Step 1: We are looking for the amount that the husband and wife each earn.

Let [latex]h=[/latex] the amount the husband earns.

Let [latex]w=[/latex] the amount the wife earns.

Step 2: Translate.

A married couple together earns [latex]$110,000[/latex].

[latex]w+h=110,000[/latex]

The wife earns [latex]$16,000[/latex] less than twice what the husband earns.

[latex]w=2h-16,000[/latex]

Step 3: The system of equations is:

[latex]\left\{\begin{array}{l}w+h=110,000\\w=2h-16,000\end{array}\right.[/latex]

Step 1: Read the problem.
Step 2: Identify what we are looking for.

We are looking for the number of calories burned each minute on the elliptical trainer and each minute of circuit training.

Step 3: Name what we are looking for.

Let [latex]e=[/latex] number of calories burned per minute on the elliptical trainer.

Let [latex]c=[/latex] number of calories burned per minute while circuit training.

Step 4: Translate into a system of equations.

10 minutes on the elliptical and circuit training for 20 minutes, burned 278 calories

[latex]10e+20c=278[/latex]

20 minutes on the elliptical and 30 minutes of circuit training burned 473 calories

[latex]20e+30c=473[/latex]

The system is:

[latex]\left\{\begin{array}{l}10e+20c=278\\20e+30c=473\end{array}\right.[/latex]

Step 5: Solve the system of equations.

Multiply the first equation by [latex]−2[/latex] to get opposite coefficients of [latex]e[/latex].

[latex]\left\{\begin{array}{l}{\color{red}{-}}{\color{red}{2}}(10e+20c)={\color{red}{-}}{\color{red}{2}}(278)\\20e+30c=473\end{array}\right.[/latex]

Simplify and add the equations.

Solve for [latex]c[/latex].

[latex]\begin{eqnarray*}\left\{\begin{array}{l}-20e-40c=-556\\20e+30c=473\end{array}\right.\\\hline-10c&=&-83\\c&=&8.3\end{eqnarray*}[/latex]

Substitute [latex]c = 8.3[/latex] into one of the original equations to solve for [latex]e[/latex].

[latex]\begin{eqnarray*}10e+20{\color{red}{c}}&=&278\\10e+20({\color{red}{8}}{\color{red}{.}}{\color{red}{3}})&=&278\\10e+166&=&278\\10e&=&112\\e&=&11.2\\\end{eqnarray*}[/latex]

Step 6: Check the answer in the problem.

Check the math on your own.

[latex]\left\{\begin{array}{l}10(11.2)+20(8.3)\overset?=278\\20(11.2)+30(8.3)\overset?=473\end{array}\right.[/latex]

Step 7: Answer the question.

Jenna burns 8.3 calories per minute circuit training and 11.2 calories per minute while on the elliptical trainer.

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for the measure of each angle.

Step 3: Name what we are looking for.

Let [latex]x=[/latex] the measure of the first angle.

Let [latex]y=[/latex] the measure of the second angle

Step 4: Translate into a system of equations.

The angles are supplementary.

[latex]x+y=180[/latex]

The larger angle is twelve less than five times the smaller angle

[latex]y=5x-12[/latex]

The system is:

[latex]\left\{\begin{array}{l}x+y=180\\y=5x-12\end{array}\right.[/latex]

Step 5: Solve the system of equations substitution.

Substitute [latex]5x-12[/latex] for [latex]y[/latex] in the first equation.

[latex]{x+}{\color{red}{5}}{{\color{red}{x}}{\color{red}{-}}}{\color{red}{12}}{=180}[/latex]

Solve for [latex]x[/latex].

[latex]\begin{eqnarray*}6x-12&=&180\\6x&=&192\\\text{Substitute in x=32}.\;\;\;\;\;\;\;y&=&5{\color{red}{x}}-12\\\text{Solve for y.}\;\;\;\;\;\;\;y&=&5\times{\color{red}{32}}-12\\y&=&160-12\\y&=&148\end{eqnarray*}[/latex]

Step 6: Check the answer in the problem.

[latex]\begin{eqnarray*}32+158&=&180✓\\5\times32-12&=&147✓\end{eqnarray*}[/latex]

Step 7: Answer the question.

The angle measures are [latex]148[/latex] and [latex]32[/latex].

Read the problem.

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

Step 1: Identify what we are looking for.

We are looking for the speed of the jet in still air and the speed of the wind.

Step 2: Name what we are looking for.

Let [latex]j=[/latex] the speed of the jet in still air.

Let [latex]w=[/latex] the speed of the wind

A chart will help us organize the information. The jet makes two trips-one in a tailwind and one in a headwind. In a tailwind, the wind helps the jet and so the rate is [latex]j+w[/latex]. In a headwind, the wind slows the jet and so the rate is [latex]j−w[/latex].

Each trip takes [latex]3[/latex] hours. In a tailwind the jet flies [latex]1095[/latex] miles. In a headwind the jet flies [latex]987[/latex] miles.

Step 3: Translate into a system of equations.

Since rate times time is distance, we get the system of equations.

[latex]\left\{\begin{array}{l}3(j+w)=1095\\3(j-w)=987\end{array}\right.[/latex]

Step 4: Solve the system of equations.

Distribute, then solve by elimination.

[latex]\begin{array}{r}\left\{\begin{array}{l}3j+3w=1095\\3j-3w=987\end{array}\right.\\\hline6j=2082\end{array}[/latex]

Add, and solve for [latex]j[/latex].

Substitute [latex]j=347[/latex] into one of the original equations, then solve for [latex]w[/latex].

[latex]\begin{eqnarray*}\text{Substitute j=347.}\;\;\;\;\;3({\color{red}{j}}+w)&=&1095\\3({\color{red}{347}}+w)&=&1095\\\text{Solve for w.}\;\;\;\;\;1041+3w&=&1095\\3w&=&54\\w&=&18\end{eqnarray*}[/latex]

Step 5: Check the answer in the problem.

With the tailwind, the actual rate of the jet would be [latex]347 + 18 = 365\;mph.[/latex]

In 3 hours the jet would travel [latex]365\times3 = 1095\;miles.[/latex]

Going into the headwind, the jet’s actual rate would be [latex]347 − 18 = 329\;mph.[/latex]

In 3 hours the jet would travel [latex]329\times3 = 987\;miles.[/latex]

Step 6: Answer the question.

The rate of the jet is [latex]347[/latex] mph and the rate of the wind is [latex]18\;mph.[/latex]