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Chapter 3: Graphing

3.4 Graphing Linear Equations

There are two common procedures that are used to draw the line represented by a linear equation. The first one is called the slope-intercept method and involves using the slope and intercept given in the equation.

If the equation is given in the form [latex]y = mx + b[/latex], then [latex]m[/latex] gives the rise over run value and the value [latex]b[/latex] gives the point where the line crosses the [latex]y[/latex]-axis, also known as the [latex]y[/latex]-intercept.

Example 3.4.1

Given the following equations, identify the slope and the [latex]y[/latex]-intercept.

• [latex]\begin{array}{lll} y = 2x - 3\hspace{0.14in} & \text{Slope }(m)=2\hspace{0.1in}&y\text{-intercept } (b)=-3 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{1}{2}x - 1\hspace{0.08in} & \text{Slope }(m)=\dfrac{1}{2}\hspace{0.1in}&y\text{-intercept } (b)=-1 \end{array}[/latex]
• [latex]\begin{array}{lll} y = -3x + 4 & \text{Slope }(m)=-3 &y\text{-intercept } (b)=4 \end{array}[/latex]
• [latex]\begin{array}{lll} y = \dfrac{2}{3}x\hspace{0.34in} & \text{Slope }(m)=\dfrac{2}{3}\hspace{0.1in} &y\text{-intercept } (b)=0 \end{array}[/latex]

When graphing a linear equation using the slope-intercept method, start by using the value given for the [latex]y[/latex]-intercept. After this point is marked, then identify other points using the slope.

This is shown in the following example.

Example 3.4.2

Graph the equation [latex]y = 2x - 3[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex]y = -3[/latex], which is placed on the coordinate [latex](0, -3).[/latex]

Now, place the next dot using the slope of 2.

A slope of 2 means that the line rises 2 for every 1 across.

Simply, [latex]m = 2[/latex] is the same as [latex]m = \dfrac{2}{1}[/latex], where [latex]\Delta y = 2[/latex] and [latex]\Delta x = 1[/latex].

Placing these points on the graph becomes a simple counting exercise, which is done as follows:

Once several dots have been drawn, draw a line through them, like so:

Note that dots can also be drawn in the reverse of what has been drawn here.

Slope is 2 when rise over run is [latex]\dfrac{2}{1}[/latex] or [latex]\dfrac{-2}{-1}[/latex], which would be drawn as follows:

Example 3.4.3

Graph the equation [latex]y = \dfrac{2}{3}x[/latex].

First, place a dot on the [latex]y[/latex]-intercept, [latex](0, 0)[/latex].

Now, place the dots according to the slope, [latex]\dfrac{2}{3}[/latex].

This will generate the following set of dots on the graph. All that remains is to draw a line through the dots.

The second method of drawing lines represented by linear equations and functions is to identify the two intercepts of the linear equation. Specifically, find [latex]x[/latex] when [latex]y = 0[/latex] and find [latex]y[/latex] when [latex]x = 0[/latex].

Example 3.4.4

Graph the equation [latex]2x + y = 6[/latex].

To find the first coordinate, choose [latex]x = 0[/latex].

This yields:

[latex]\begin{array}{lllll} 2(0)&+&y&=&6 \\ &&y&=&6 \end{array}[/latex]

Coordinate is [latex](0, 6)[/latex].

Now choose [latex]y = 0[/latex].

[latex]\begin{array}{llrll} 2x&+&0&=&6 \\ &&2x&=&6 \\ &&x&=&\frac{6}{2} \text{ or } 3 \end{array}[/latex]

Coordinate is [latex](3, 0)[/latex].

Draw these coordinates on the graph and draw a line through them.

Example 3.4.5

Graph the equation [latex]x + 2y = 4[/latex].

[latex]\begin{array}{llrll} (0)&+&2y&=&4 \\ &&y&=&\frac{4}{2} \text{ or } 2 \end{array}[/latex]

Coordinate is [latex](0, 2)[/latex].

[latex]\begin{array}{llrll} x&+&2(0)&=&4 \\ &&x&=&4 \end{array}[/latex]

Coordinate is [latex](4, 0)[/latex].

Example 3.4.6

Graph the equation [latex]2x + y = 0[/latex].

[latex]\begin{array}{llrll} 2(0)&+&y&=&0 \\ &&y&=&0 \end{array}[/latex]

Coordinate is [latex](0, 0)[/latex].

Since the intercept is [latex](0, 0)[/latex], finding the other intercept yields the same coordinate. In this case, choose any value of convenience.

Choose [latex]x = 2[/latex].

[latex]\begin{array}{rlrlr} 2(2)&+&y&=&0 \\ 4&+&y&=&0 \\ -4&&&&-4 \\ \hline &&y&=&-4 \end{array}[/latex]

Coordinate is [latex](2, -4)[/latex].

For questions 1 to 10, sketch each linear equation using the slope-intercept method.

• [latex]y = -\dfrac{1}{4}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x - 4[/latex]
• [latex]y = -\dfrac{3}{5}x + 1[/latex]
• [latex]y = -\dfrac{4}{3}x + 2[/latex]
• [latex]y = \dfrac{5}{3}x + 4[/latex]
• [latex]y = \dfrac{3}{2}x - 5[/latex]
• [latex]y = -\dfrac{2}{3}x - 2[/latex]
• [latex]y = -\dfrac{4}{5}x - 3[/latex]
• [latex]y = \dfrac{1}{2}x[/latex]

For questions 11 to 20, sketch each linear equation using the [latex]x\text{-}[/latex] and [latex]y[/latex]-intercepts.

• [latex]x + 4y = -4[/latex]
• [latex]2x - y = 2[/latex]
• [latex]2x + y = 4[/latex]
• [latex]3x + 4y = 12[/latex]
• [latex]4x + 3y = -12[/latex]
• [latex]x + y = -5[/latex]
• [latex]3x + 2y = 6[/latex]
• [latex]x - y = -2[/latex]
• [latex]4x - y = -4[/latex]

For questions 21 to 28, sketch each linear equation using any method.

• [latex]y = -\dfrac{1}{2}x + 3[/latex]
• [latex]y = 2x - 1[/latex]
• [latex]y = -\dfrac{5}{4}x[/latex]
• [latex]y = -3x + 2[/latex]
• [latex]y = -\dfrac{3}{2}x + 1[/latex]
• [latex]y = \dfrac{1}{3}x - 3[/latex]
• [latex]y = \dfrac{3}{2}x + 2[/latex]
• [latex]y = 2x - 2[/latex]

For questions 29 to 40, reduce and sketch each linear equation using any method.

• [latex]y + 3 = -\dfrac{4}{5}x + 3[/latex]
• [latex]y - 4 = \dfrac{1}{2}x[/latex]
• [latex]x + 5y = -3 + 2y[/latex]
• [latex]3x - y = 4 + x - 2y[/latex]
• [latex]4x + 3y = 5 (x + y)[/latex]
• [latex]3x + 4y = 12 - 2y[/latex]
• [latex]2x - y = 2 - y \text{ (tricky)}[/latex]
• [latex]7x + 3y = 2(2x + 2y) + 6[/latex]
• [latex]x + y = -2x + 3[/latex]
• [latex]3x + 4y = 3y + 6[/latex]
• [latex]2(x + y) = -3(x + y) + 5[/latex]
• [latex]9x - y = 4x + 5[/latex]

Answer Key 3.4

Intermediate Algebra Copyright © 2020 by Terrance Berg is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License , except where otherwise noted.

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Key to Algebra Workbooks

Key to Algebra offers a unique, proven way to introduce algebra to your students. New concepts are explained in simple language, and examples are easy to follow. Word problems relate algebra to familiar situations, helping students to understand abstract concepts. Students develop understanding by solving equations and inequalities intuitively before formal solutions are introduced. Students begin their study of algebra in Books 1-4 using only integers. Books 5-7 introduce rational numbers and expressions. Books 8-10 extend coverage to the real number system.

Linear Equations

A linear equation is an equation for a straight line

These are all linear equations:

Let us look more closely at one example:

Example: y = 2x + 1 is a linear equation:

The graph of y = 2x+1 is a straight line

• When x increases, y increases twice as fast , so we need 2x
• When x is 0, y is already 1. So +1 is also needed
• And so: y = 2x + 1

Here are some example values:

Check for yourself that those points are part of the line above!

Different Forms

There are many ways of writing linear equations, but they usually have constants (like "2" or "c") and must have simple variables (like "x" or "y").

Examples: These are linear equations:

But the variables (like "x" or "y") in Linear Equations do NOT have:

• Exponents (like the 2 in x 2 )
• Square roots , cube roots , etc

Examples: These are NOT linear equations:

Slope-intercept form.

The most common form is the slope-intercept equation of a straight line :

Example: y = 2x + 1

• Slope: m = 2
• Intercept: b = 1

Point-Slope Form

Another common one is the Point-Slope Form of the equation of a straight line:

Example: y − 3 = (¼)(x − 2)

It is in the form y − y 1 = m(x − x 1 ) where:

General Form

And there is also the General Form of the equation of a straight line:

Example: 3x + 2y − 4 = 0

It is in the form Ax + By + C = 0 where:

There are other, less common forms as well.

As a Function

Sometimes a linear equation is written as a function , with f(x) instead of y :

And functions are not always written using f(x):

The Identity Function

There is a special linear function called the "Identity Function":

And here is its graph:

It is called "Identity" because what comes out is identical to what goes in:

Constant Functions

Another special type of linear function is the Constant Function ... it is a horizontal line:

No matter what value of "x", f(x) is always equal to some constant value.

Using Linear Equations

You may like to read some of the things you can do with lines:

• Finding the Midpoint of a Line Segment
• Finding Parallel and Perpendicular Lines
• Finding the Equation of a Line from 2 Points

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Unit 3: Linear relationships

Lesson 3: representing proportional relationships.

• Graphing proportional relationships: unit rate (Opens a modal)
• Graphing proportional relationships from a table (Opens a modal)
• Graphing proportional relationships from an equation (Opens a modal)
• Graphing proportional relationships Get 3 of 4 questions to level up!

Lesson 4: Comparing proportional relationships

• Rates & proportional relationships example (Opens a modal)
• Rates & proportional relationships: gas mileage (Opens a modal)
• Rates & proportional relationships Get 5 of 7 questions to level up!

Lesson 7: Representations of linear relationships

• Linear & nonlinear functions: missing value (Opens a modal)

Lesson 8: Translating to y=mx+b

• Intro to slope-intercept form (Opens a modal)
• Graph from slope-intercept equation (Opens a modal)

Lesson 9: Slopes don't have to be positive

• Intro to intercepts (Opens a modal)
• Slope-intercept equation from slope & point (Opens a modal)
• Linear & nonlinear functions: word problem (Opens a modal)
• Intercepts from a graph Get 3 of 4 questions to level up!
• Slope from graph Get 3 of 4 questions to level up!
• Slope-intercept intro Get 3 of 4 questions to level up!
• Graph from slope-intercept form Get 3 of 4 questions to level up!
• Slope-intercept equation from graph Get 3 of 4 questions to level up!

Lesson 10: Calculating slope

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• Slope from two points Get 3 of 4 questions to level up!

Lesson 11: Equations of all kinds of lines

• Converting to slope-intercept form (Opens a modal)

Extra practice: Slope

• Intro to slope (Opens a modal)
• Worked examples: slope-intercept intro (Opens a modal)
• Graphing slope-intercept form (Opens a modal)
• Writing slope-intercept equations (Opens a modal)
• Slope-intercept form review (Opens a modal)
• Slope-intercept from two points Get 3 of 4 questions to level up!

Lesson 12: Solutions to linear equations

• Solutions to 2-variable equations (Opens a modal)
• Worked example: solutions to 2-variable equations (Opens a modal)
• Solutions to 2-variable equations Get 3 of 4 questions to level up!

Lesson 13: More solutions to linear equations

• Completing solutions to 2-variable equations (Opens a modal)
• Complete solutions to 2-variable equations Get 3 of 4 questions to level up!

Extra practice: Intercepts

• x-intercept of a line (Opens a modal)
• Intercepts from an equation (Opens a modal)
• Worked example: intercepts from an equation (Opens a modal)
• Intercepts of lines review (x-intercepts and y-intercepts) (Opens a modal)
• Intercepts from an equation Get 3 of 4 questions to level up!

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3.2: Solve Applications with Systems of Equations

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Learning Objectives

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

• Solve direct translation applications
• Solve geometry applications

Solve uniform motion applications

Before you get started, take this readiness quiz.

• The sum of twice a number and nine is 31. Find the number. If you missed this problem, review [link] .
• Twins Jon and Ron together earned $96,000 last year. Ron earned $8000 more than three times what Jon earned. How much did each of the twins earn? If you missed this problem, review [link] .
• An express train and a local train leave Pittsburgh to travel to Washington, D.C. The express train can make the trip in four hours and the local train takes five hours for the trip. The speed of the express train is 12 miles per hour faster than the speed of the local train. Find the speed of both trains. If you missed this problem, review [link] .

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.

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.

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

Example \(\PageIndex{2}\)

The sum of two numbers is 10. One number is 4 less than the other. Find the numbers.

Example \(\PageIndex{3}\)

The sum of two numbers is \(−6\). One number is 10 less than the other. Find the numbers.

\(2, −8\)

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?

Example \(\PageIndex{5}\)

Geraldine has been offered positions by two insurance companies. The first company pays a salary of $12,000 plus a commission of $100 for each policy sold. The second pays a salary of $20,000 plus a commission of $50 for each policy sold. How many policies would need to be sold to make the total pay the same?

160 policies

Example \(\PageIndex{6}\)

Kenneth currently sells suits for company A at a salary of $22,000 plus a $10 commission for each suit sold. Company B offers him a position with a salary of $28,000 plus a $4 commission for each suit sold. How many suits would Kenneth need to sell for the options to be equal?

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

Example \(\PageIndex{7}\)

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?

Example \(\PageIndex{8}\)

Mark went to the gym and did 40 minutes of Bikram hot yoga and 10 minutes of jumping jacks. He burned 510 calories. The next time he went to the gym, he did 30 minutes of Bikram hot yoga and 20 minutes of jumping jacks burning 470 calories. How many calories were burned for each minute of yoga? How many calories were burned for each minute of jumping jacks?

Mark burned 11 calories for each minute of yoga and 7 calories for each minute of jumping jacks.

Example \(\PageIndex{9}\)

Erin spent 30 minutes on the rowing machine and 20 minutes lifting weights at the gym and burned 430 calories. During her next visit to the gym she spent 50 minutes on the rowing machine and 10 minutes lifting weights and burned 600 calories. How many calories did she burn for each minutes on the rowing machine? How many calories did she burn for each minute of weight lifting?

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

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 \(\PageIndex{10}\)

Translate to a system of equations and then solve.

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

\(\begin{array} {ll} {\textbf{Step 1. Read }\text{the problem.}} &{} \\ {\textbf{Step 2. Identify }\text{what we are looking for.}} &{\text{We are looking for the measure of each}} \\ {} &{\text{angle.}} \\ {\textbf{Step 3. Name }\text{what we are looking for.}} &{\text{Let} x=\text{ the measure of the first angle.}} \\ {} &{\hspace{3mm} y= \text{ the measure of the second angle}} \\ {\textbf{Step 4. Translate }\text{into a system of}} &{\text{The angles are complementary.}} \\ {\text{equations.}} &{\hspace{15mm} x+y=90} \\ {} &{\text{The difference of the two angles is 26}} \\ {} &{\text{degrees.}} \\ {} &{\hspace{15mm} x−y=26} \\ {} &{} \\ {} &{} \\ {\text{The system is shown.}} &{\hspace{15mm} \left\{ \begin{array} {l} x+y=90 \\ x−y=26 \end{array} \right. } \\ {} &{} \\ {} &{} \\ {\textbf{Step 5. Solve }\text{the system of equations} } &{\hspace{15mm} \left\{ \begin{array} {l} x+y=90 \\ \underline{x−y=26} \end{array} \right. } \\ {\text{by elimination.}} &{\hspace{21mm} 2x\hspace{4mm}=116} \\ {} &{\hspace{28mm} x=58} \\ {} &{} \\ {} &{} \\ {\text{Substitute }x=58\text{ into the first equation.}} &{\hspace{15mm} x+y=90} \\ {} &{\hspace{14mm} 58+y=90} \\ {} &{\hspace{22mm} y=32} \\ {\textbf{Step 6. Check }\text{the answer in the problem.}} &{} \\ {} &{} \\ {} &{} \\ {} &{} \\ {\hspace{15mm} 58+32=90\checkmark} &{} \\ {\hspace{15mm} 58−32=26\checkmark} &{} \\ {\textbf{Step 7. Answer }\text{the question.}} &{\text{The angle measures are 58 and 32 degrees.}} \end{array} \)

Example \(\PageIndex{11}\)

The difference of two complementary angles is 20 degrees. Find the measures of the angles.

The angle measures are 55 and 35.

Example \(\PageIndex{12}\)

The difference of two complementary angles is 80 degrees. Find the measures of the angles.

The angle measures are 5 and 85.

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

Example \(\PageIndex{13}\)

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.

Example \(\PageIndex{14}\)

Two angles are supplementary. The measure of the larger angle is 12 degrees more than three times the smaller angle. Find the measures of the angles.

The angle measures are 42 and 138.

Example \(\PageIndex{15}\)

Two angles are supplementary. The measure of the larger angle is 18 less than twice the measure of the smaller angle. Find the measures of the angles.

The angle measures are 66 and 114.

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 \(\PageIndex{16}\)

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.

We will draw and label a figure.

Example \(\PageIndex{17}\)

The measure of one of the small angles of a right triangle is 2 more than 3 times the measure of the other small angle. Find the measure of both angles.

Example \(\PageIndex{18}\)

The measure of one of the small angles of a right triangle is 18 less than twice the measure of the other small angle. Find the measure of both angles.

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

Example \(\PageIndex{19}\)

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.

Example \(\PageIndex{20}\)

Mario wants to put a 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 155 feet of fencing to enclose the pool. The length of the long side is 10 feet less than twice the width. Find the length and width of the pool area to be enclosed.

The length is 60 feet and the width is 35 feet.

Example \(\PageIndex{21}\)

Alexis wants to build a rectangular dog run in her yard adjacent to her neighbor’s fence. She will use 136 feet of fencing to completely enclose the rectangular dog run. The length of the dog run along the neighbor’s fence will be 16 feet less than twice the width. Find the length and width of the dog run.

The length is 60 feet and the width is 38 feet.

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 \(\PageIndex{22}\)

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?

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.

Since \(D=r·t\) we can fill in the Distance column.

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

\(\hspace{85mm} 65j=78k \nonumber \)

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

\( \hspace{105mm} k=j-\frac{1}{2} \nonumber \)

\(\begin{array} {ll} {\text{Now we have the system.}} &{\left\{ \begin{array} {l} k=j−\frac{1}{2} \\ 65j=78k \end{array} \right.} \\ {\textbf{Solve }\text{the system of equations by substitution.}} &{} \\ {} &{} \\ {\text{Substitute }k=j−12\text{ into the second equation,}} &{} \\ {\text{then solve for }j.} &{} \\ {} &{65j=78k} \\ {} &{65j=78(j−\frac{1}{2})} \\ {} &{65j=78j−39} \\ {} &{−13j=−39} \\ {} &{j=3} \\{\begin{array} {l} {\text{To find Kelly’s time, substitute }j=3 \text{ into the first}} \\ {\text{equation, then solve for }k.} \end{array} } &{k=j−\frac{1}{2}} \\ {} &{k=3−\frac{1}{2} } \\ {} &{k=\frac{5}{2} \text{ or } k=2\frac{1}{2}} \\ {\textbf{Check }\text{the answer in the problem.}} &{} \\ {\begin{array} {lllll} {\text{Joni}} &{3 \text{ hours}} &{(65\text{ mph})} &= &{195\text{ miles}} \\ {\text{Kelly}} &{2\frac{1}{2} \text{ hours}} &{(78\text{ mph})} &= &{195\text{ miles}} \end{array}} &{} \\ {\text{Yes, they will have traveled the same distance}} &{} \\{\text{when they meet.}} &{} \\ {\textbf{Answer }\text{the question.}} &{} \\ {} &{\text{Kelly will catch up to Joni in}} \\ {} &{2\frac{1}{2}\text{ hours. By then, Joni will}} \\ {} &{\text{have traveled }3 \text{ hours.}} \\ \end{array}\)

Example \(\PageIndex{23}\)

Mitchell left Detroit on the interstate driving south towards Orlando at a speed of 60 miles per hour. Clark left Detroit 1 hour later traveling at a speed of 75 miles per hour, following the same route as Mitchell. How long will it take Clark to catch Mitchell?

It will take Clark 4 hours to catch Mitchell.

Example \(\PageIndex{24}\)

Charlie left his mother’s house traveling at an average speed of 36 miles per hour. His sister Sally left 15 minutes \((\frac{1}{4} \text{ hour})\) later traveling the same route at an average speed of 42 miles per hour. How long before Sally catches up to Charlie?

It will take Sally \(112\) 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.

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 is \(b+c\).

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 \(\PageIndex{25}\)

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.

Example \(\PageIndex{26}\)

A Mississippi river boat cruise sailed 120 miles upstream for 12 hours and then took 10 hours to return to the dock. Find the speed of the river boat in still water and the speed of the river current.

The rate of the boat is 11 mph and the rate of the current is 1 mph.

Example \(\PageIndex{27}\)

Jason paddled his canoe 24 miles upstream for 4 hours. It took him 3 hours to paddle back. Find the speed of the canoe in still water and the speed of the river current.

The speed of the canoe is 7 mph and the speed of the current is 1 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 \(\PageIndex{28}\)

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.

Example \(\PageIndex{29}\)

A small jet can fly 1,325 miles in 5 hours with a tailwind but only 1,035 miles in 5 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 235 mph and the speed of the wind is 30 mph.

Example \(\PageIndex{30}\)

A commercial jet can fly 1,728 miles in 4 hours with a tailwind but only 1,536 miles in 4 hours into a headwind. Find the speed of the jet in still air and the speed of the wind.

The speed of the jet is 408 mph and the speed of the wind is 24 mph.

Access this online resource for additional instruction and practice with systems of equations.

• Systems of Equations

Key Concepts