4.4 Solve Applications with Systems of Equations

Learning Objectives

By the end of this section, you will be able to:

  • Translate to a system of equations
  • Solve direct translation applications
  • Solve geometry applications
  • Solve uniform motion applications

Try It

Before you get started, take this readiness quiz:

1) The sum of twice a number and nine is [latex]31[/latex]. Find the number.
2) Twins Jon and Ron together earned [latex]$96,000[/latex] last year. Ron earned [latex]$8,000[/latex] more than three times what Jon earned. How much did each of the twins earn?
3) Alessio rides his bike [latex]3\frac{1}{2}[/latex] hours at a rate of [latex]10[/latex] miles per hour. How far did he ride?

Previously in this chapter, we solved several applications with systems of linear equations. In this section, we’ll look at some specific types of applications that relate to two quantities. We’ll translate the words into linear equations, decide which is the most convenient method to use, and then solve them.

We will use our Problem-Solving Strategy for Systems of Linear Equations.

How to

Use a problem-solving strategy for systems of linear equations.

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

 

Translate to a System of Equations

Many of the problems we solved in earlier applications related to two quantities. Here are two of the examples from the chapter on Math Models.

  • The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.
  • A married couple together earns $110,000 a year. The wife earns $16,000 less than twice what her husband earns. What does the husband earn?

In that chapter, we translated each situation into one equation using only one variable. Sometimes it was a bit of a challenge figuring out how to name the two quantities, wasn’t it?

Let’s see how we can translate these two problems into a system of equations with two variables. We’ll focus on Steps 1 through 4 of our Problem-Solving Strategy.

Example 1

How to Translate to a System of Equations

Translate to a system of equations:

The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.

Solution
This figure has four rows and three columns. The first row reads, “Step 1: Read the problem. Make sure you understand all the words and ideas. This is a number problem. The sum of two numbers is negative fourteen. One number is four less than the other. Find the numbers.”
Figure 4.4.1
The second row reads, “Step 2: Identify what you are looking for. ‘Find the numbers.’ We are looking for 2 numbers.”
Figure 4.4.2
The third row reads, “Step 3: Name what you are looking for. Choose variables to represent those quantities. We will use two variables, m and n. Let me = one number n = second number.”
Figure 4.4.3
The fourth row reads, “Step 4: Translate into a system of equations. We will write one equation for each sentence.” The figure then shows how, “The sum of the numbers is -14” becomes m + n = -14 and “One number is four less than the other” becomes m = n – 4. The figure then says, “The system is m + n = -14 and m = n – 4.”
Figure 4.4.4

Try It

4) Translate to a system of equations: The sum of two numbers is negative twenty-three. One number is 7 less than the other. Find the numbers.

Solution

[latex]\left\{\begin{array}{l}m+n=-23\\m=n-7\end{array}\right.[/latex]

5) Translate to a system of equations: The sum of two numbers is negative eighteen. One number is 40 more than the other. Find the numbers.

Solution

[latex]\left\{\begin{array}{l}m+n=-18\\m=n+40\end{array}\right.[/latex]

We’ll do another example where we stop after we write the system of equations.

Example 2

Translate to a system of equations:

A married couple together earns [latex]$110,000[/latex] a year. The wife earns [latex]$16,000[/latex] less than twice what her husband earns. What does the husband earn?

Solution

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]

Try It

6) Translate to a system of equations: A couple has a total household income of $84,000. The husband earns $18,000 less than twice what the wife earns. How much does the wife earn?

Solution

[latex]\left\{\begin{array}{l}w+h=84,000\\h=2w-18,000\end{array}\right.[/latex]

7) Translate to a system of equations: A senior employee makes $5 less than twice what a new employee makes per hour. Together they make $43 per hour. How much does each employee make per hour?

Solution

[latex]\left\{\begin{array}{l}s=2n-5\\s+n=43\end{array}\right.[/latex]

Solve Direct Translation Applications

We set up but did not solve, the systems of equations in Examples 1 and 2. Now we’ll translate a situation to a system of equations and then solve it.

Example 3

Translate to a system of equations and then solve: Devon is 26 years older than his son Cooper. The sum of their ages is 50. Find their ages.

Solution

Step 1: Read the problem.

Step 2: Identify what we are looking for.

We are looking for the ages of Devon and Cooper.

Step 3: Name what we are looking for.

Let [latex]d=[/latex] Devon’s age.

Let [latex]c=[/latex] Cooper’s age

Step 4: Translate into a system of equations.

Devon is 26 years older than Cooper.

[latex]d=c+26[/latex]

The sum of their ages is 50.

[latex]d+c=50[/latex]

The system is:

[latex]\left\{\begin{array}{l}d=c+26\\d+c=50\end{array}\right.[/latex]

Step 5: Solve the system of equations.

Solve by substitution.

d=c+26 d+c=50 d+c=50
Figure 4.4.5

Substitute [latex]c + 26[/latex] into the second equation.

[latex]{\color{red}{c}}{{\color{red}{+}}}{\color{red}{26}}{+c=50}[/latex]

Solve for [latex]c[/latex].

[latex]\begin{eqnarray*}2c+26&=&50\\2c&=&24\\{\color{red}{c}}\;&=&\;{\color{red}{12}}\\\text{Substitute c = 12.}\;\;\;\;\;\;\;d&=&{\color{red}{c}}+26\\\text{Solve for d.}\;\;\;\;\;\;\;d&=&{\color{red}{12}}+26\\d&=&38\\\end{eqnarray*}[/latex]

Step 6: Check the answer in the problem.

Is Devon’s age [latex]26[/latex] more than Cooper’s?
Yes, [latex]38[/latex] is [latex]26[/latex] more than [latex]12[/latex].
Is the sum of their ages [latex]50[/latex]?
Yes, [latex]38[/latex] plus [latex]12[/latex] is [latex]50[/latex].

Step 7: Answer the question.

Devon is [latex]38[/latex] and Cooper is 12 years old.

Try It

8) Translate to a system of equations and then solve: Ali is [latex]12[/latex] years older than his youngest sister, Jameela. The sum of their ages is [latex]40[/latex]. Find their ages.

Solution

Ali is [latex]26[/latex] and Jameela is [latex]14[/latex].

9) Translate to a system of equations and then solve: Jake’s dad is [latex]6[/latex] more than [latex]3[/latex] times Jake’s age. The sum of their ages is [latex]42[/latex]. Find their ages.

Solution

Jake is [latex]9[/latex] and his dad is [latex]33[/latex].

Example 4

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 said she burned 278 calories. When she spent 20 minutes on the elliptical trainer and 30 minutes on circuit training she burned 473 calories. How many calories does she burn for each minute on the elliptical trainer? How many calories does she burn 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 [latex]e=[/latex] be the number of calories burned per minute on the elliptical trainer.

Let [latex]c=[/latex] be 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

[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 to 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.

Try It

10) Translate to a system of equations and then solve: Mark went to the gym and did [latex]40[/latex] minutes of Bikram hot yoga and [latex]10[/latex] minutes of jumping jacks. He burned [latex]510[/latex] calories. The next time he went to the gym, he did [latex]30[/latex] minutes of Bikram hot yoga and [latex]20[/latex] minutes of jumping jacks burning [latex]470[/latex] calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Solution

Mark burned [latex]11[/latex] calories for each minute of yoga and [latex]7[/latex] calories for each minute of jumping jacks.

11) Translate to a system of equations and then solve: Erin spent [latex]30[/latex] minutes on the rowing machine and [latex]20[/latex] minutes lifting weights at the gym and burned [latex]430[/latex] calories. During her next visit to the gym, she spent [latex]50[/latex] minutes on the rowing machine and [latex]10[/latex] minutes lifting weights and burned [latex]600[/latex] calories. How many calories did she burn for each minute on the rowing machine? How many calories did she burn for each minute of weight lifting?

Solution

Erin burned [latex]11[/latex] calories for each minute on the rowing machine and [latex]5[/latex] calories for each minute of weight lifting.

Solve Geometry Applications

When we learned about Math Models, we solved geometry applications using properties of triangles and rectangles. Now we’ll add 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 [latex]90[/latex] degrees.
Two angles are supplementary if the sum of the measures of their angles is [latex]180[/latex] 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 5

Translate to a system of equations and then solve: The difference of two complementary angles is [latex]26[/latex] 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 [latex]x=[/latex] the measure of the first angle.

Let [latex]m=[/latex]the measure of the second angle.

Step 4: Translate into a system of equations.

The angles are complementary.

[latex]x+y=90[/latex]

The difference of the two angles is 26 degrees.

[latex]x-y=26[/latex]

The system is

[latex]\left\{\begin{array}{l}x+y=90\\x-y=26\end{array}\right.[/latex]

Step 5: Solve the system of equations by elimination.

[latex]\left\{\begin{array}{l}x+y=90\\x-y=26\end{array}\right.[/latex]

[latex]\begin{eqnarray*}2x&=&116\\x&=&58\end{eqnarray*}[/latex]

Substitute [latex]x=58[/latex] into the first equation.

[latex]\begin{eqnarray*}x+y&=&90\\58+y&=&90\\y&=&32\end{eqnarray*}[/latex]

Step 6: Check the answer in the problem.

[latex]\begin{eqnarray*}58+32&=&90✓\\58-32&=&26✓\end{eqnarray*}[/latex]

Step 7: Answer the question.

The angle measures are [latex]58[/latex] degrees and [latex]32[/latex] degrees.

Try It 

12) Translate to a system of equations and then solve: The difference between two complementary angles is [latex]20[/latex] degrees. Find the measures of the angles.

Solution

The angle measures are [latex]55[/latex] degrees and [latex]35[/latex] degrees.

13) Translate to a system of equations and then solve: The difference between two complementary angles is [latex]80[/latex] degrees. Find the measures of the angles.

Solution

The angle measures are [latex]5[/latex] degrees and [latex]85[/latex] degrees.

Example 6

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 [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.

x+y=180 y=5x-12 x+y=180
Figure 4.4.6

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 to 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].

Try It 

14) Translate to a system of equations and then solve: Two angles are supplementary. The measure of the larger angle is [latex]12[/latex] degrees more than three times the smaller angle. Find the measures of the angles.

Solution

The angle measures are [latex]42[/latex] degrees and [latex]138[/latex] degrees.

15) Translate to a system of equations and then solve: Two angles are supplementary. The measure of the larger angle is [latex]18[/latex] less than twice the measure of the smaller angle. Find the measures of the angles.

Solution

The angle measures are [latex]66[/latex] degrees and [latex]114[/latex] degrees.

Example 7

Translate to a system of equations and then solve:

Randall has [latex]125[/latex] feet of fencing to enclose the rectangular 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 [latex]5[/latex] 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 you are looking for.

We are looking for the length and width.

Diagram of a rectangular fence labelled L and W
Figure 4.4.7

Step 3: Name what we are looking for.

Let [latex]L=[/latex] the length of the fenced yard.

Let [latex]W=[/latex]the width of the fenced yard

Step 4: Translate into a system of equations.

One length and two widths equal [latex]125[/latex].

[latex]L+2W=125[/latex]

The length will be [latex]5[/latex] feet more than four times the width.

[latex]L=4W+5[/latex]

The system is:

[latex]\left\{\begin{array}{l}L+2W=125\\L=4W+5\end{array}\right.[/latex]

Step 5: Solve the system of equations by substitution.

L+2W=125 L=4W+5 L+2W=125
Figure 4.4.8

Substitute [latex]L = 4W + 5[/latex] into the first equation, then solve for [latex]W[/latex].

[latex]\begin{eqnarray*}{\color{red}{4}}{\color{red}{W}}{\color{red}{+}}{\color{red}{5}}+2W&=&125\\6W+5&=&125\\6W&=&120\end{eqnarray*}[/latex]

Substitute [latex]20[/latex] for [latex]W[/latex] in the second equation, then solve for [latex]L[/latex].

[latex]\begin{eqnarray*}\text{Substitute W=20.}\;\;\;\;\;\;L&=&4{\color{red}{W}}+5\\\text{Solve for L.}\;\;\;\;\;\;L&=&4\times{\color{red}{20}}+5\\L&=&80+5\\L&=&85\end{eqnarray*}[/latex]

Step 6: Check the answer in the problem.

[latex]\begin{eqnarray*}20+28+20&=&125✓\\85&=&4\times20+5✓\end{eqnarray*}[/latex]

Step 7: Answer the equation.

The length is [latex]85[/latex] feet and the width is [latex]20[/latex] feet.

Try It 

16) Translate to a system of equations and then solve: Mario wants to put a rectangular fence around the pool in his backyard. Since one side is adjacent to the house, he will only need to fence three sides. There are two long sides and the one shorter side is parallel to the house. He needs [latex]155[/latex] feet of fencing to enclose the pool. The length of the long side is [latex]10[/latex] feet less than twice the width. Find the length and width of the pool area to be enclosed.

Solution

The length is [latex]60[/latex] feet and the width is [latex]35[/latex] feet.

17) Translate to a system of equations and then solve: Alexis wants to build a rectangular dog run in her yard adjacent to her neighbour’s fence. She will use [latex]136[/latex] feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbour’s fence will be [latex]16[/latex] feet less than twice the width. Find the length and width of the dog run.

Solution

The length is [latex]60[/latex] feet and the width is [latex]38[/latex] feet.

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 [latex]D=rt[/latex] where [latex]D[/latex] is the distance travelled, [latex]r[/latex] is the rate, and [latex]t[/latex] 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 [latex]65[/latex] miles per hour. Half an hour later, Kelly left St. Louis on the same route as Joni, driving [latex]78[/latex] 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.

This figure shows a diagram. Denver is on the left and St. Louis is on the right. There is a ray stretching from St. Louis to Denver. It is labeled “Joni” and “65 m p h.” There is another ray stretching from St. Louis to Denver. It is labeled “Kelly (1/2 hour later)” and “78 m p h.”
Figure 4.4.9

Step 1: 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, [latex]k[/latex], and Joni, [latex]j[/latex], will each drive. Since [latex]D=r\cdot t[/latex] we can fill in the Distance column.

Type Rate x Time = Distance
Joni 65 j 65j
Kelly 78 k 78k

Step 2: 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, [latex]65j=78k[/latex].

Also, since Kelly left later, her time will be [latex]\frac{1}{2}[/latex] hour less than Joni’s time.

So, [latex]k=j-\frac{1}{2}[/latex]

Now we have the system.

[latex]\left\{\begin{array}{l}k=j-\frac12\\65j=78k\end{array}\right.[/latex]

Step 3: Solve the system of equations by substitution.

[latex]65j=78k[/latex]

Substitute [latex]k=j-\frac{1}{2}[/latex] into the second equation, then solve for [latex]j[/latex].

[latex]\begin{eqnarray*}65j&=&78(j-\frac12)\\65j&=&78j-39\\-13j&=&-39\\{\color{red}{j}}&=&{\color{red}{3}}\end{eqnarray*}[/latex]

To find Kelly’s time, substitute [latex]j=3[/latex] into the first equation, then solve for [latex]k[/latex].

[latex]\begin{eqnarray*}\text{Solve for k.}\;k&=&{\color{red}{j}}-\frac12\\k&=&{\color{red}{3}}-\frac12\\k&=&\frac52\;or\;k&=&2\frac12\end{eqnarray*}[/latex]

Step 4: Check the answer to the problem.

Joni [latex]3[/latex] hours ([latex]65[/latex]mph) = [latex]195[/latex] miles.

Kelly [latex]2\frac{1}{2}[/latex] hours ([latex]78[/latex] mph) = [latex]195[/latex] miles.

Yes, they will have travelled the same distance when they meet.

Step 5: Answer the question.

Kelly will catch up to Joni in [latex]2\frac{1}{2}[/latex] hours. By then, Joni will have travelled [latex]3[/latex] hours.

Try It 

18) Translate to a system of equations and then solve: Mitchell left Detroit on the interstate driving south towards Orlando at a speed of [latex]60[/latex] miles per hour. Clark left Detroit [latex]1[/latex] hour later travelling at a speed of [latex]75[/latex] miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

Solution

It will take Clark [latex]4[/latex] hours to catch Mitchell.

19) Translate to a system of equations and then solve: Charlie left his mother’s house travelling at an average speed of [latex]36[/latex] miles per hour. His sister Sally left [latex]15[/latex] minutes ([latex]\frac{1}{4}[/latex] hour) later travelling the same route at an average speed of [latex]42[/latex] miles per hour. How long before Sally catches up to Charlie?

Solution

It will take Sally [latex]1\frac{1}{2}[/latex] hours to catch up to Charlie.

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.

Figures 4.4.1 and 4.4.2 show how a river current affects the speed at which a boat is travelling. We’ll call the speed of the boat in still water [latex]b[/latex] and the speed of the river current [latex]c[/latex].

In Figure 4.4.1 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 is [latex]b+c[/latex].

This figure shows a boat floating in water. On the right, there is an arrow pointing towards the boat. It is labeled “c.” On the left, there is an arrow pointing away from the boat. It is labeled “b.”
Figure 4.4.10

In Figure 4.4.2 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 [latex]b-c[/latex].

This figure shows a boat floating in water. To the left is an arrow pointing away from the boat labeled “b,” and an arrow pointing towards the boat labeled “c.”
Figure 4.4.11

We’ll put some numbers to this situation in Example 9

Example 9

Translate to a system of equations and then solve:

A river cruise ship sailed [latex]60[/latex] miles downstream for [latex]4[/latex] hours and then took [latex]5[/latex] 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.

This figure shows an arrow labeled “c” which continues to the right, representing the wave. Under the wave is a ray that points to the right and is labeled “four hours.” Under this ray is another ray pointing to the left labeled “five hours.” It is the same length as the ray labeled “four hours.” There is a bracket under the ray labeled “five hours.” The bracket is labeled “60 miles.”
Figure 4.4.12

Step 1: Identify what we are looking for.

We are looking for the speed of the ship in still water and the speed of the current.

Step 2: Name what we are looking for.

Let [latex]s=[/latex] the rate of the ship in still water.

Let [latex]c=[/latex] the rate of the current

A chart will help us organize the information. The ship goes downstream and then upstream. Going downstream, the current helps the ship; therefore, the ship’s actual rate is [latex]s+c[/latex]. Going upstream, the current slows the ship; therefore, the actual rate is [latex]s−c[/latex].

Type Rate x Time = Distance
Downstream [latex]s+c[/latex] 4 60
Upstream [latex]s-c[/latex] 5 60

Downstream it takes [latex]4[/latex] hours. Upstream it takes [latex]5[/latex] hours. Each way the distance is [latex]60[/latex] miles.

Step 3: Translate into a system of equations.

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

[latex]\left\{\begin{array}{l}4(s+c)=60\\5(s-c)=60\end{array}\right.[/latex]

Step 4: Solve the system of equations.

Distribute to put both equations in standard form, then solve by elimination.

[latex]\left\{\begin{array}{l}4s+4c=60\\5s-5c=60\end{array}\right.[/latex]

Multiply the top equation by [latex]5[/latex] and the bottom equation by [latex]4[/latex]. Add the equations, then solve for [latex]s[/latex].

[latex]\begin{array}{r}\left\{\begin{array}{l}20s+20c=300\\20s-20c=240\end{array}\right.\\\hline40s=540\end{array}[/latex]

Substitute [latex]s=13.5[/latex] into one of the original equations.

[latex]\begin{eqnarray*}\text{Substitute s=13.5.}\;\;\;\;\;4({\color{red}{s}}+c)&=&60\\4({\color{red}{13}}{\color{red}{.}}{\color{red}{5}}+c)&=&60\\\text{Solve for c.}\;\;\;\;\;54+4c&=&60\\4c&=&6\\c&=&1.5\end{eqnarray*}[/latex]

Step 5: Check the answer in the problem.

The downstream rate would be [latex]13.5 + 1.5 = 15\;mph[/latex].

In 4 hours the ship would travel [latex]15\cdot 4 = 60\;miles[/latex].

The upstream rate would be [latex]13.5 − 1.5 = 12\;mph[/latex].

In 5 hours the ship would travel [latex]12 \cdot 5 = 60\;miles[/latex].

Step 6: Answer the question.

The rate of the ship is [latex]13.5[/latex] mph and the rate of the current is [latex]1.5[/latex] mph.

Try It 

20) Translate to a system of equations and then solve A Mississippi river boat cruise sailed [latex]120[/latex] miles upstream for [latex]12[/latex] hours and then took [latex]10[/latex] hours to return to the dock. Find the speed of the riverboat in still water and the speed of the river current.

Solution

The rate of the boat is [latex]11[/latex] mph and the rate of the current is [latex]1[/latex] mph.

21) Translate to a system of equations and then solve: Jason paddled his canoe [latex]24[/latex] miles upstream for [latex]4[/latex] hours. It took him [latex]3[/latex] hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

Solution

The speed of the canoe is [latex]7[/latex] mph and the speed of the current is [latex]1[/latex] mph.

Wind currents affect airplane speeds in the same way as water currents affect boat speeds. We’ll see this in Example 10. 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 [latex]1095[/latex] miles in three hours with a tailwind but only [latex]987[/latex] 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.

This figure shows an arrow labeled “3 hours” which continues to the right, representing the wind. Under the wave is a ray that points to the right and is labeled “j plus w equals 365” and “1,095 miles”. Under this ray is another ray pointing to the left labeled “j minus w equals 329” and “987 miles.”
4.4.13

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].

Type Rate x Time = Distance
Tailwind [latex]j+w[/latex] 3 1095
Headwind [latex]j-w[/latex] 3 987

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]

Try It

22) Translate to a system of equations and then solve: A small jet can fly [latex]1,325[/latex] miles in [latex]5[/latex] hours with a tailwind but only [latex]1025[/latex] miles in [latex]5[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The speed of the jet is [latex]235[/latex] mph and the speed of the wind is [latex]30[/latex] mph.

23) Translate to a system of equations and then solve: A commercial jet can fly [latex]1728[/latex] miles in [latex]4[/latex] hours with a tailwind but only [latex]1536[/latex] miles in [latex]4[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Solution

The speed of the jet is [latex]408[/latex] mph and the speed of the wind is [latex]24[/latex] mph.

Glossary

complementary angles
Two angles are complementary if the sum of the measures of their angles is 90 degrees.
supplementary angles
Two angles are supplementary if the sum of the measures of their angles is 180 degrees.

Exercises: Translate to a System of Equations

Instructions: For questions 1-14, translate to a system of equations and solve the system.

1. The sum of two numbers is fifteen. One number is three less than the other. Find the numbers.
Solution

The numbers are [latex]6[/latex] and [latex]9[/latex].


2. The sum of two numbers is twenty-five. One number is five less than the other. Find the numbers.

3. The sum of two numbers is negative thirty. One number is five times the other. Find the numbers.
Solution

The numbers are [latex]-5[/latex] and [latex]-25[/latex].


4. The sum of two numbers is negative sixteen. One number is seven times the other. Find the numbers.

5. Twice a number plus three times a second number is twenty-two. Three times the first number plus four times the second is thirty-one. Find the numbers.
Solution

The numbers are [latex]5[/latex] and [latex]4[/latex].


6. Six times a number plus twice a second number is four. Twice the first number plus four times the second number is eighteen. Find the numbers.

7. Three times a number plus three times a second number is fifteen. Four times the first plus twice the second number is fourteen. Find the numbers.
Solution

The numbers are [latex]2[/latex] and [latex]3[/latex].


8. Twice a number plus three times a second number is a negative one. The first number plus four times the second number is two. Find the numbers.

9. A married couple together earn [latex]$75\text{,}000[/latex]. The husband earns [latex]$15\text{,}000[/latex] more than five times what his wife earns. What does the wife earn?
Solution

[latex]$10\text{,}000[/latex]


10. During two years in college, a student earned [latex]$9\text{,}500[/latex]. The second year she earned [latex]$500[/latex] more than twice the amount she earned the first year. How much did she earn the first year?

11. Daniela invested a total of [latex]$50\text{,}000[/latex], some in a certificate of deposit (CD) and the remainder in bonds. The amount invested in bonds was [latex]$5000[/latex] more than twice the amount she put into the CD. How much did she invest in each account?
Solution

She put [latex]$15\text{,}000[/latex] into a CD and [latex]$35\text{,}000[/latex] in bonds.


12. Jorge invested [latex]$28\text{,}000[/latex] into two accounts. The amount he put in his money market account was [latex]$2\text{,}000[/latex] less than twice what he put into a CD. How much did he invest in each account?

13. In her last two years in college, Marlene received [latex]$42\text{,}000[/latex] in loans. The first year she received a loan that was [latex]$6\text{,}000[/latex] less than three times the amount of the second year’s loan. What was the amount of her loan for each year?
Solution

The amount of the first year’s loan was [latex]$30\text{,}000[/latex] and the amount of the second year’s loan was [latex]$12\text{,}000[/latex].


14. Jen and David owe [latex]$22\text{,}000[/latex] in loans for their two cars. The amount of the loan for Jen’s car is [latex]$2000[/latex] less than twice the amount of the loan for David’s car. How much is each car loan?

Exercises: Solve Direct Translation Applications

Instructions: For questions 15-24, translate to a system of equations and solve.

15. Alyssa is twelve years older than her sister, Bethany. The sum of their ages is forty-four. Find their ages.
Solution

Bethany is [latex]16[/latex] years old and Alyssa is [latex]28[/latex] years old.


16. Robert is [latex]15[/latex] years older than his sister, Helen. The sum of their ages is sixty-three. Find their ages.

17. The age of Noelle’s dad is six less than three times Noelle’s age. The sum of their ages is seventy-four. Find their ages.
Solution

Noelle is [latex]20[/latex] years old and her dad is [latex]54[/latex] years old.


18. The age of Mark’s dad is [latex]4[/latex] less than twice Marks’s age. The sum of their ages is ninety-five. Find their ages.

19. Two containers of gasoline hold a total of fifty gallons. The big container can hold ten gallons less than twice the small container. How many gallons does each container hold?
Solution

The small container holds [latex]20[/latex] gallons and the large container holds [latex]30[/latex] gallons.


20. June needs [latex]48[/latex] gallons of punch for a party and has two different coolers to carry it in. The bigger cooler is five times as large as the smaller cooler. How many gallons can each cooler hold?

21. Shelly spent [latex]10[/latex] minutes jogging and [latex]20[/latex] minutes cycling and burned [latex]300[/latex] calories. The next day, Shelly swapped times, doing [latex]20[/latex] minutes of jogging and [latex]10[/latex] minutes of cycling and burned the same number of calories. How many calories were burned for each minute of jogging and how many for each minute of cycling?
Solution

There were [latex]10[/latex] calories burned jogging and [latex]10[/latex] calories burned cycling.


22. Drew burned [latex]1800[/latex] calories Friday playing one hour of basketball and canoeing for two hours. Saturday he spent two hours playing basketball and three hours canoeing and burned [latex]3200[/latex] calories. How many calories did he burn per hour when playing basketball?

23. Troy and Lisa were shopping for school supplies. Each purchased different quantities of the same notebook and thumb drive. Troy bought four notebooks and five thumb drives for [latex]$116[/latex]. Lisa bought two notebooks and three thumb dives for [latex]$68[/latex]. Find the cost of each notebook and each thumb drive.
Solution

Notebooks are [latex]$4[/latex] and thumb drives are [latex]$20[/latex].


24. Nancy bought seven pounds of oranges and three pounds of bananas for [latex]$17[/latex]. Her husband later bought three pounds of oranges and six pounds of bananas for [latex]$12[/latex]. What was the cost per pound of the oranges and the bananas?

Exercises: Solve Geometry Applications

Instructions: For questions 25-40, translate to a system of equations and solve.

25. The difference of two complementary angles is [latex]30[/latex] degrees. Find the measures of the angles.
Solution

The measures are [latex]60[/latex] degrees and [latex]30[/latex] degrees.


26. The difference of two complementary angles is [latex]68[/latex] degrees. Find the measures of the angles.

27. The difference of two supplementary angles is [latex]70[/latex] degrees. Find the measures of the angles.
Solution

The measures are [latex]125[/latex] degrees and [latex]55[/latex] degrees.


28. The difference of two supplementary angles is [latex]24[/latex] degrees. Find the measure of the angles.

29. The difference of two supplementary angles is [latex]8[/latex] degrees. Find the measures of the angles.
Solution

[latex]94[/latex] degrees and [latex]86[/latex] degrees


30. The difference of two supplementary angles is [latex]88[/latex] degrees. Find the measures of the angles.

31. The difference of two complementary angles is [latex]55[/latex] degrees. Find the measures of the angles.
Solution

[latex]72.5[/latex] degrees and [latex]17.5[/latex] degrees


32. The difference of two complementary angles is [latex]17[/latex] degrees. Find the measures of the angles.

33. Two angles are supplementary. The measure of the larger angle is four more than three times the measure of the smaller angle. Find the measures of both angles.
Solution

The measures are [latex]44[/latex] degrees and [latex]136[/latex] degrees.


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

35. Two angles are complementary. The measure of the larger angle is twelve less than twice the measure of the smaller angle. Find the measures of both angles.
Solution

The measures are [latex]34[/latex] degrees and [latex]56[/latex] degrees.


36. Two angles are complementary. The measure of the larger angle is ten more than four times the measure of the smaller angle. Find the measures of both angles.

37. Wayne is hanging a string of lights [latex]45[/latex] feet long around the three sides of his rectangular patio, which is adjacent to his house. The length of his patio, the side along the house, is five feet longer than twice its width. Find the length and width of the patio.
Solution

The width is [latex]10[/latex] feet and the length is [latex]25[/latex] feet.


38. Darrin is hanging [latex]200[/latex] feet of Christmas garland on the three sides of fencing that enclose his rectangular front yard. The length, the side along the house, is five feet less than three times the width. Find the length and width of the fencing.

39. A frame around a rectangular family portrait has a perimeter of [latex]60[/latex] inches. The length is fifteen less than twice the width. Find the length and width of the frame.
Solution

The width is [latex]15[/latex] feet and the length is [latex]15[/latex] feet.


40. The perimeter of a rectangular toddler play area is [latex]100[/latex] feet. The length is ten more than three times the width. Find the length and width of the play area.

Exercises: Solve Uniform Motion Applications

Instructions: For questions 41-52, translate to a system of equations and solve.

41. Sarah left Minneapolis heading east on the interstate at a speed of [latex]60[/latex] mph. Her sister followed her on the same route, leaving two hours later and driving at a rate of [latex]70[/latex] mph. How long will it take for Sarah’s sister to catch up to Sarah?

Solution

It took Sarah’s sister [latex]12[/latex] hours.


42. College roommates John and David were driving home to the same town for the holidays. John drove [latex]55[/latex] mph, and David, who left an hour later, drove [latex]60[/latex] mph. How long will it take for David to catch up to John?

43. At the end of spring break, Lucy left the beach and drove back towards home, driving at a rate of [latex]40[/latex] mph. Lucy’s friend left the beach for home [latex]30[/latex] minutes (half an hour) later, and drove [latex]50[/latex] mph. How long did it take Lucy’s friend to catch up to Lucy?
Solution

It took Lucy’s friend [latex]2[/latex] hours.


44. Felecia left her home to visit her daughter driving [latex]45[/latex] mph. Her husband waited for the dog sitter to arrive and left home twenty minutes ([latex]\frac13[/latex] hour) later. He drove [latex]55[/latex] mph to catch up to Felecia. How long before he reaches her?

45. The Jones family took a [latex]12[/latex] mile canoe ride down the Indian River in two hours. After lunch, the return trip back up the river took three hours. Find the rate of the canoe in still water and the rate of the current.
Solution

The canoe rate is [latex]5[/latex] mph and the current rate is [latex]1[/latex] mph.


46. A motor boat travels [latex]60[/latex] miles down a river in three hours but takes five hours to return upstream. Find the rate of the boat in still water and the rate of the current.

47. A motor boat travelled [latex]18[/latex] miles down a river in two hours but going back upstream, it took [latex]4.5[/latex] hours due to the current. Find the rate of the motor boat in still water and the rate of the current. (Round to the nearest hundredth.).
Solution

The boat rate is [latex]6.5[/latex] mph and the current rate is [latex]2.5[/latex] mph.


48. A river cruise boat sailed [latex]80[/latex] miles down the Mississippi River for four hours. It took five hours to return. Find the rate of the cruise boat in still water and the rate of the current. (Round to the nearest hundredth.).

49. A small jet can fly [latex]1\text{,}072[/latex] miles in [latex]4[/latex] hours with a tailwind but only [latex]848[/latex] miles in [latex]4[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
Solution

The jet rate is [latex]240[/latex] mph and the wind speed is [latex]28[/latex] mph.


50. A small jet can fly [latex]1\text{,}435[/latex] miles in [latex]5[/latex] hours with a tailwind but only [latex]1\text{,}215[/latex] miles in [latex]5[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

51. A commercial jet can fly [latex]868[/latex] miles in [latex]2[/latex] hours with a tailwind but only [latex]792[/latex] miles in [latex]2[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.
Solution

The jet rate is [latex]415[/latex] mph and the wind speed is [latex]19[/latex] mph.


52. A commercial jet can fly [latex]1\text{,}320[/latex] miles in [latex]3[/latex] hours with a tailwind but only [latex]1\text{,}170[/latex] miles in [latex]3[/latex] hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

Exercises: Everyday Math

Instructions: For questions 53-54, answer the given everyday math word problems.

53. At a school concert, [latex]425[/latex] tickets were sold. Student tickets cost [latex]$5[/latex] each and adult tickets cost [latex]$8[/latex] each. The total receipts for the concert were [latex]$2\text{,}851[/latex].

Solve the system[latex]\left\{\begin{array}{c}s+a=425\\ 5s+8a=2\text{,}851\end{array}\right.[/latex] to find [latex]s[/latex], the number of student tickets and [latex]a[/latex], the number of adult tickets.

Solution

[latex]s=183,a=242[/latex]


54. The first graders at one school went on a field trip to the zoo. The total number of children and adults who went on the field trip was [latex]115[/latex]. The number of adults was [latex]\frac{1}{4}[/latex] the number of children.

Solve the system [latex]\left\{\begin{array}{c}c+a=115\\ a=\frac{1}{4}c\end{array}\right.[/latex] to find [latex]c[/latex], the number of children and [latex]a[/latex], the number of adults.

Exercises: Writing Exercises

Instructions: For questions 55-56, answer the given writing exercises.

55. Write an application problem similar to (Example 4.4.3) using the ages of two of your friends or family members. Then translate to a system of equations and solve it.
Solution

Answers will vary.


56. Write a uniform motion problem similar to (Example 4.4.8) that relates to where you live with your friends or family members. Then translate to a system of equations and solve it.
definition

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