**4.2 Solve Applications with Systems of Equations**

Topics covered in this section are:

**4.2.1 Solve Direct Translation Applications**

Systems of linear equations are very useful for solving applications. Some people find setting up word problems with two variables easier than setting them up with just one variable. To solve an application, we’ll first translate the words into a system of linear equations. Then we will decide the most convenient method to use, and then solve the system.

**HOW TO: Solve applications with systems of equations.**

**Read**the problem. Make sure all the words and ideas are understood.**Identify**what we are looking for.**Name**what we are looking for. Choose variables to represent those quantities.**Translate**into a system of equations.**Solve**the system of equations using good algebra techniques.**Check**the answer in the problem and make sure it makes sense.**Answer**the question with a complete sentence.

We solved number problems with one variable earlier. Let’s see how differently it works using two variables.

**Example 1 **

The sum of two numbers is zero. One number is nine less than the other. Find the numbers.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what we are looking for. | We are looking for two numbers. |

Step 3. Name what we are looking for. | Let $n=$the first number and $m=$ the second number. |

Step 4. Translate into a system of equations. | The sum of two numbers is zero. $n+m=0$ One number is nine less than the other. $n=m-9$ |

The system is: | $\Bigg\{ \begin{align*} n+m&=0 \\ n&=m-9 \end{align*}$ |

Step 5. Solve the system of equations. We will use substitution since the second equation is solved for $n$. | |

Substitute $m-9$ for $n$ in the first equation and solve for $m$. | $n+m=0$ $\textcolor{red}{m-9}+m=0$ $2m-9=0$ $2m=9$ $m=\frac{9}{2}$ |

Substitute $m=\frac{9}{2}$ into the second equation and then solve for $n$. | $n=m-9$ $n=\textcolor{red}{\frac{9}{2}}-9$ $n=\frac{9}{2}-\frac{18}{2}$ $n=-\frac{9}{2}$ |

Step 6. Check the answer in the problem. | Do these numbers make sense in the problem? We will leave this to you. |

Step 7. Answer the question. | The numbers are $\frac{9}{2}$ and $-\frac{9}{2}$. |

**Example 2 **

Heather has been offered two options for her salary as a trainer at the gym. Option A would pay her $\$25,000$ plus $\$15$ for each training session. Option B would pay her $\$10,000+\$40$ for each training session. How many training sessions would make the salary options equal?

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what we are looking for. | We are looking for the number of training sessions that would make the pay equal. |

Step 3. Name what we are looking for. | Let $s=$ Heather’s salary and $n=$ the number of training sessions. |

Step 4. Translate into a system of equations. | Option A would pay her $\$25,000$ plus $\$15$ for each training session. $s=25000+15n$ Option B would pay her $\$10,000$ plus $\$40$ for each training session. $s=10000+40n$ |

The system is: | $\Bigg\{ \begin{align*} s&=25000+15n \\ s&=10000+40n \end{align*}$ |

Step 5. Solve the system of equations. We will use substitution. | |

Substitute $25000+15n$ for $s$ in the second equation and solve for $n$. | $s=10000+40n$ $\textcolor{red}{25000+15n}=10000+40n$ $15000=25n$ $600=n$ |

Step 6. Check the answer. | Are $600$ training sessions a year reasonable? Are the two options equal when $n=600$? |

Step 7. Answer the question. | The salary options would be equal for $600$ training sessions. |

As you solve each application, remember to analyze which method of solving the system of equations would be most convenient.

**Example 3 **

Translate to a system of equations and then solve:

When Jenna spent $10$ minutes on the elliptical trainer and then did circuit training for $20$ minutes, her fitness app says she burned $278$ calories. When she spent $20$ minutes on the elliptical trainer and $30$ minutes circuit training she burned $473$ calories. How many calories does she burn for each minute on the elliptical trainer? How many calories for each minute of circuit training?

**Solution**

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 $e=$ the number of calories burned per minute on the elliptical trainer and $c=$ the 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. $10e+20c=278$ $20$ minutes on the elliptical and $30$ minutes of circuit training burned $473$ calories. $20e+30c=473$ |

The system is: | $\Bigg\{ \begin{align*} 10e+20c&=278 \\ 20e+30c&=473 \end{align*}$ |

Step 5. Solve the system of equations. We will use elimination. | |

Multiply the first equation by $-2$ to get opposite coefficients. | $\Bigg\{ \begin{align*} \textcolor{red}{-2}(10e+20c)&=\textcolor{red}{-2}(278) \\ 20e+30c&=473 \end{align*}$ |

Simplify and add the equations. Solve for $c$. | $\Bigg\{ \begin{align*} -20e-40c&=-556 \\ 20e+30c&=473 \end{align*}$ $\overline{ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -10c=-83}$ $c=8.3$ |

Substitute $c=8.3$ into one of the original equations to solve for $e$. | $\begin{align*} 10e+20c&=278 \\ 10e+20(\textcolor{red}{8.3})&=278 \\ 10e+166&=278 \\ 10e&=112 \\ e&=11.2 \end{align*}$ |

Step 6. Check the answer. | Check the math on you own. $10(11.2)+20(8.3)=278$ ? $20(11.2)+30(8.3)=478$ ? |

Step 7. Answer the question. | Jenna burns $8.3$ calories per minute circuit training and $11.2$ calories per minute while on the elliptical. |

**4.2.2 Solve Geometry Applications**

We will now solve geometry applications using systems of linear equations. We will need to add complementary angles and supplementary angles to our list some properties of angles.

The measures of two **complementary angles** add to $90$ degrees. The measures of two **supplementary angles** add to $180$ degrees.

**COMPLEMENTARY AND SUPPLEMENTARY ANGLES**

Two angles are complementary if the sum of the measures of their angles is $90$ degrees.

Two angles are supplementary if the sum of the measures of their angles is $180$ degrees.

If two angles are complementary, we say that *one angle is the complement of the other.*

If two angles are supplementary, we say that *one angle is the supplement of the other.*

**Example 4**

Translate to a system of equations and then solve.

The difference of two complementary angles is $26$ degrees. Find the measures of the angles.

**Solution**

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 $x=$ the measure of the first angle and $y=$ the measure of the second angle. |

Step 4. Translate into a system of equations. | The angles are complementary. $x+y=90$ The difference of the two angles is $26$ degrees. $x-y=26$ |

The system is shown. | $\Bigg\{ \begin{align*} x+y&=90 \\ x-y&=26 \end{align*}$ |

Step 5. Solve the system of equations by elimination. | $\Bigg\{ \begin{align*} x+y&=90 \\ x-y&=26 \end{align*}$ $ \ \ \overline{\ \ \ \ 2x \ \ \ \ \ =116}$ $x=58$ |

Substitute $x=58$ into the first equation. | $\begin{align*} x+y=90 \\ 58+y=90 \\ y=32 \end{align*}$ |

Step 6. Check the answer. | $58+32=90 \ \checkmark$ $58-32=26 \ \checkmark$ |

Step 7. Answer the question. | The angle measures are $58$ and $32$ degrees. |

In the next example, we remember that the measures of supplementary angles add to $180$.

**Example 5**

Translate to a system of equations and then solve:

Two angles are supplementary. The measure of the larger angle is twelve degrees less than five times the measure of the smaller angle. Find the measures of both angles.

**Solution**

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 $x=$ the measure of the first angle and $y=$ the measure of the second angle. |

Step 4. Translate into a system of equations. | The angles are supplementary $x+y=180$ The larger angle is twelve less than 5 times the smaller angle. $y=5x-12$ |

The system is shown. | $\Bigg\{ \begin{align*} x+y&=180 \\ y&=5x-12 \end{align*}$ |

Step 5. Solve the system of equations by substitution. | |

Substitute $5x-12$ in for $y$ in the first equation. Solve for $x$. | $\begin{align*} x+y&=180 \\ x+\textcolor{red}{5x-12}&=180 \\ 6x&=192\\ x&=32 \end{align*}$ |

Substitute $32$ in for $x$ in the second equation, then solve for $y$. | $\begin{align*} y&=5x-12 \\ y&=5\cdot \textcolor{red}{32}-12 \\ y&=160-12 \\ y&=148 \end{align*}$ |

Step 6. Check the answer. | $32+148=180 \ \checkmark$ $5\cdot 32-12=148 \ \checkmark$ |

Step 7. Answer the question. | The angle measures are $148$ and $32$ degrees. |

Recall that the angles of a triangle add up to $180$ degrees. A right triangle has one angle that is $90$ degrees. What does that tell us about the other two angles? In the next example we will be finding the measures of the other two angles.

**Example 6**

The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. Find the measures of both angles.

**Solution**

We will draw and label a figure.

Step 1. Read the problem. | |

Step 2. Identify what we are looking for. | We are looking for the measures of the angles. |

Step 3. Name what we are looking for. | Let $a=$ the measure of the first angle and $b=$ the measure of the second angle. |

Step 4. Translate into a system of equations. | The measure of one of the small angles of a right triangle is ten more than three times the measure of the other small angle. $a=3b+10$ The sum of the measures of the angles of a triangle is $180$. $a+b+90=180$ |

The system is shown. | $\Bigg\{ \begin{align*} a&=3b+10 \\ a+b+90&=180 \end{align*}$ |

Step 5. Solve the system of equations. We will use substitution since the first equation is solve for $a$. | |

Substitute $3b+10$ in for $a$ in the second equation. Solve for $b$. | $\begin{align*} a+b+90&=180 \\ (\textcolor{red}{3b+10})+b+90&=180 \\ 4b+100&=180\\ 4b&=80\\ b&=20\end{align*}$ |

Substitute $b=20$ into the first equation, then solve for $a$. | $\begin{align*} a&=3b+10 \\ a&=3\cdot \textcolor{red}{20}+10 \\ a&=70 \end{align*}$ |

Step 6. Check the answer. | We will leave this to you. |

Step 7. Answer the question. | The measures of the small angles are are $20$ and $70$ degrees. |

Often it is helpful when solving geometry applications to draw a picture to visualize the situation.

**Example 7**

Translate to a system of equations and then solve:

Randall has $125$ feet of fencing to enclose the part of his backyard adjacent to his house. He will only need to fence around three sides, because the fourth side will be the wall of the house. He wants the length of the fenced yard (parallel to the house wall) to be $5$ feet more than four times as long as the width. Find the length and the width.

**Solution**

Step 1. Read the problem. | |

Step 2. Identify what we are looking for. | We are looking for the length and width. |

Step 3. Name what we are looking for. | Let $L=$ the length of the fenced yard and $W=$ the width of the fenced yard. |

Step 4. Translate into a system of equations. | One length and two widths equal $125$. $L+2W=125$ The length will be $5$ feet more than four times the width. $L=4W+5$ |

The system is shown. | $\Bigg\{ \begin{align*} L+2W&=125 \\ L&=4W+5 \end{align*}$ |

Step 5. Solve the system of equations by substitution. | |

Substitute $L=4W+5$ into the first equation, then solve for $W$. | $\begin{align*} L+2W&=125 \\ (\textcolor{red}{4W+5})+2W&=125 \\ 6W+5&=125\\ 6W&=120\\ W&=20\end{align*}$ |

Substitute $20$ for $W$ in the second equation, then solve for $L$. | $\begin{align*} L&=4W+5 \\ L&=4\cdot \textcolor{red}{20}+5 \\ L&=80+5\\ L&=85 \end{align*}$ |

Step 6. Check the answer in the problem. | $(85)+2(2)=125 \ \checkmark$ $85=4(20)+5 \ \checkmark$ |

Step 7. Answer the question. | The length is $85$ feet and the width is $20$ feet. |

**4.2.3 Solve uniform motion applications**

We used a table to organize the information in uniform motion problems when we introduced them earlier. We’ll continue using the table here. The basic equation was $D=rt$ where $D$ is the distance traveled, $r$ is the rate, and $t$ is the time.

Our first example of a uniform motion application will be for a situation similar to some we have already seen, but now we can use two variables and two equations.

**Example 8 **

Translate to a system of equations and then solve:

Joni left St. Louis on the interstate, driving west towards Denver at a speed of $65$ miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving $78$ miles per hour. How long will it take Kelly to catch up to Joni?

**Solution**

A diagram is useful in helping us visualize the situation.

**Identify and name** what we are looking for. A chart will help us organize the data. We know the rates of both Joni and Kelly, and so we enter them in the chart. We are looking for the length of time Kelly, $k$, and Joni, $j$, will each drive.

Rate $\cdot$ | Time $=$ | Distance | |

Joni | $65$ | $j$ | $65j$ |

Kelly | $78$ | $k$ | $78k$ |

Since $D=r \cdot t$ we can fill in the Distance column.

**Translate** into a system of equations.

To make the system of equations, we must recognize that Kelly and Joni will drive the same distance. So,

$65j=78k$

Also, since Kelly left later, her time will be $\frac{1}{2}$ hour less than Joni’s time. So,

$k=j-\frac{1}{2}$ | |

Now we have the system. | $\Bigg\{ \begin{align*} k&=j-\frac{1}{2} \\ 65j&=78k \end{align*}$ |

Solve the system of equations by substitution. | |

Substitute $k=j-\frac{1}{2}$ into the second equation, then solve for $j$. | $\begin{align*} 65j&=78k \\ 65j&=78\Big(j-\frac{1}{2}\Big) \\ 65j&=78j-39 \\ -13j&=-39 \\ j&=3 \end{align*}$ |

To find Kelly’s time, substitute $j=3$ into the first equation, then solve for $k$. | $\begin{align*} k&=j-\frac{1}{2} \\ k&=3-\frac{1}{2} \end{align*}$ $k=\frac{5}{2}$ or $k=2\frac{1}{2}$ |

Check the answer in the problem. | Joni $3$ hours ($65$ mph) $=195$ miles Kelly $2\frac{1}{2}$ hours ($78$ mph) $=195$ miles Yes, they will have traveled the same distance when they meet. |

Answer the question. | Kelly will catch up to Joni in $2\frac{1}{2}$ hours. By then, Joni will have traveled $3$ hours. |

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 $b$ and the speed of the river current $c$.

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 $b+c$.

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 $b−c$.

We’ll put some numbers to this situation in the next example.

**Example 9**

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.

**Solution**

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 $s=$ the rate of the ship in still water and $c=$ the rate of the current |

A chart will help us organize the information. The ship goes downstream then upstream. Going downstream, the current helps the ship, so the ship’s actual rate is $s+c$. Going upstream, the current slows the ship, so the actual rate is $s-c$. | |

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. | $\Bigg\{ \begin{align*} 4(s+c)&=60 \\ 5(s-c)&=60 \end{align*}$ |

Solve the system of equations.Distribute to put both equations in standard form, then solve by elimination. | $\Bigg\{ \begin{align*} 4s+4c&=60 \\ 5s-5c&=60 \end{align*}$ |

Multiply the top equation by $5$ and the bottom equation by $4$. Add the equations, then solve for $s$. | $\Bigg\{ \begin{align*} 20s+20c&=300 \\ 20s-20c&=240 \end{align*}$ $\overline{ \ \ \ 40s \ \ \ \ \ \ \ \ \ \ \ =540}$ $s=13.5$ |

Substitute $s=13.5$ into one of the original equations. then solve for $c$. | $\begin{align*} 4(s+c)&=60 \\ 4(\textcolor{red}{13.5}+c)&=60 \\ 54+4c&=60 \\ 4c&=6\\ c&=1.5 \end{align*}$ |

Check the answer in the problem. | The downstream rate would be $13.5+1.5=15$ mph. In $4$ hours the ship would travel $15 \cdot 4=60$ miles. The upstream rate would be $13.5-1.5=12$ mph. In $5$ hours the ship would travel $12 \cdot 5=60$ miles. |

Answer the question. | The rate of the ship is $13.5$ mph and the rate of the current is $1.5$ mph. |

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in the next example. A wind current in the same direction as the plane is flying is called a *tailwind*. A wind current blowing against the direction of the plane is called a *headwind*.

**Example 10**

Translate to a system of equations and then solve:

A private jet can fly $1,095$ miles in three hours with a tailwind but only $987$ miles in three hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

**Solution**

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

Identify what we are looking for. | We are looking for the speed of the jet in still air and the speed of the wind. |

Name what we are looking for. | Let $j=$ the speed of the jet in still air and $w=$ the speed of the wind. |

A chart will help us organize the information. The jet make two trips – one in tailwind and one in headwind. In a tailwind, the wind helps the jet, so the rate is $j+w$. In a headwind, the wind slows the jet, so the rate is $j-w$. | |

Each trip takes $3$ hours. In a tailwind the jet flies $1,095$ miles. In a headwind the jet flies $987$ miles. | |

Translate into a system of equations.Since rate times time is distance, we get the system of equations. | $\Bigg\{ \begin{align*} 3(j+w)&=1095 \\ 3(j-w)&=987 \end{align*}$ |

Solve the system of equations.Distribute, then solve by elimination. Add, and solve for $j$. | $\Bigg\{ \begin{align*} 3j+3w&=1095 \\ 3j-3w&=987 \end{align*}$ $\overline{ \ \ \ 6j \ \ \ \ \ \ \ \ \ \ \ =2082}$ $j=347$ |

Substitute $j=347$ into one of the original equations. then solve for $w$. | $\begin{align*} 3(j+w)&=1095 \\ 3(\textcolor{red}{347}+w)&=1095 \\ 1041+3w&=1095 \\ 3w&=54\\ w&=18 \end{align*}$ |

Check the answer in the problem. | With the tailwind, the actual rate of the jet would be $347+18=365$ mph. In $3$ hours the jet would travel $365 \cdot 3=1,095$ miles. Going into the headwind, the jet’s actual rate would be $347-18=329$ mph. In $3$ hours the jet would travel $329 \cdot 3=987$ miles. |

Answer the question. | The rate of the jet is $347$ mph and the rate of the wind is $18$ mph. |

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*Marecek, L., & Mathis, A. H. (2020). Solve Applications with Systems of Equations. In Intermediate Algebra 2e. OpenStax. https://openstax.org/books/intermediate-algebra-2e/pages/4-2-solve-applications-with-systems-of-equations*.*License: CC BY 4.0. Access for free at https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction*