ORCCA Linear Equations and Inequalities with Fractions

Section3.2Linear Equations and Inequalities with Fractions

ObjectivesPCC Course Content and Outcome Guide

C.1.1:4.e
C.1.1:5.a
C.1.1:4.f

In this section, we will learn how to solve linear equations and inequalities with fractions.

Figure3.2.1Alternative Video Lessons

Subsection3.2.1Introduction

So far, in our last step of solving for a variable we have divided each side of the equation by a constant, as in:

\begin{align*} 2x\amp=10\\ \divideunder{2x}{2} \amp= \divideunder{10}{2}\\ x\amp=5 \end{align*}

If we have a coefficient that is a fraction, we could proceed in exactly the same manner:

\begin{align*} \frac{1}{2}x\amp=10\\ \divideunder{\frac{1}{2}x}{\frac{1}{2}} \amp= \divideunder{10}{\frac{1}{2}}\\ x\amp=10\cdot \frac{2}{1}=20 \end{align*}

What if our equation or inequality was more complicated though, for example $\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{?}$ We would have to first do a lot of fraction arithmetic in order to then divide each side by the coefficient of $x\text{.}$ An alternate approach is to instead multiply each side of the equation by a chosen constant that eliminates the denominator. In the equation $\frac{1}{2}x=10\text{,}$ we could simply multiply each side of the equation by $2\text{,}$ which would eliminate the denominator of $2\text{:}$

\begin{align*} \frac{1}{2}x\amp=10\\ \multiplyleft{2}\left(\frac{1}{2}x\right)\amp=\multiplyleft{2}10\\ x\amp=20 \end{align*}

For more complicated equations, we will multiply each side of the equation by the least common denominator (LCD) of all fractions contained in the equation.

Subsection3.2.2Eliminating Denominators

Example3.2.2

A $4$-foot tree was planted. The tree grows $\frac{2}{3}$ of a foot every year. How many years later will the tree reach $10$ feet?

Since the tree grows $\frac{2}{3}$ of a foot every year, we can use a table to help write a formula modeling the tree's growth:

Years Passed	Tree's Height (ft)
$0$	$4$
$1$	$4+\frac{2}{3}$
$2$	$4+\frac{2}{3}\cdot2$
$\vdots$	$\vdots$
$y$	$4+\frac{2}{3}y$

From this, we've determined that $y$ years since the tree was planted, the tree's height will be $4+\frac{2}{3}y$ feet.

To find when the tree will be $10$ feet tall, we write and solve this equation:

\begin{align*} 4+\frac{2}{3}y\amp=10\\ \multiplyleft{3}\left(4+\frac{2}{3}y\right)\amp=\multiplyleft{3}10\\ 3\cdot4+3\cdot\frac{2}{3}y\amp=30\\ 12+2y\amp=30\\ 2y\amp=18\\ y\amp=9 \end{align*}

Now we will check the solution $9$ in the equation $4+\frac{2}{3}y=10\text{:}$

\begin{align*} 4+\frac{2}{3}y\amp=10\\ 4+\frac{2}{3}(\substitute{9})\amp\stackrel{?}{=}10\\ 4+6\amp\stackrel{\checkmark}{=}10\\ 10\amp\stackrel{\checkmark}{=}10 \end{align*}

In summary, the tree will be $10$ feet tall $9$ years later.

Let's look at a few more examples.

Example3.2.3

Solve for $x$ in $\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{.}$

Solution

To solve this equation, we first need to identify the LCD of all fractions in the equation. On the left side we have $\frac{1}{4}$ and $\frac{2}{3}\text{.}$ On the right side we have $\frac{1}{6}\text{.}$ The LCD of $3, 4\text{,}$ and $6$ is $12\text{,}$ so we will multiply each side of the equation by $12$ in order to eliminate all of the denominators:

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \multiplyleft{12}\left(\frac{1}{4}x+\frac{2}{3}\right)\amp=\multiplyleft{12}\frac{1}{6}\\ 12\cdot\left(\frac{1}{4}x\right)+12\cdot\left(\frac{2}{3}\right)\amp=12\cdot\frac{1}{6}\\ 3x+8\amp=2\\ 3x\amp=-6\\ \divideunder{3x}{3}\amp=\divideunder{-6}{3}\\ x\amp=-2 \end{align*}

Checking the solution $-2$ in the equation $\frac{1}{4}x+\frac{2}{3}=\frac{1}{6}\text{:}$

\begin{align*} \frac{1}{4}x+\frac{2}{3}\amp=\frac{1}{6}\\ \frac{1}{4}(\substitute{-2})+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{2}{4}+\frac{2}{3}\amp\stackrel{?}{=}\frac{1}{6}\\ -\frac{6}{12}+\frac{8}{12}\amp\stackrel{?}{=}\frac{1}{6}\\ \frac{2}{12}\amp\stackrel{\checkmark}{=}\frac{1}{6} \end{align*}

The solution is therefore $-2$ and the solution set is $\{-2\}\text{.}$

Example3.2.4

Solve for $z$ in $-\frac{2}{5}z-\frac{3}{2}=-\frac{1}{2}z+\frac{4}{5}\text{.}$

Solution

The first thing we need to do is identify the LCD of all denominators in this equation. Since the denominators are $2$ and $5\text{,}$ the LCD is $10\text{.}$ So as our first step, we will multiply each side of the equation by $10$ in order to eliminate all denominators:

\begin{align*} -\frac{2}{5}z-\frac{3}{2}\amp=-\frac{1}{2}z+\frac{4}{5}\\ \multiplyleft{10}\left(-\frac{2}{5}z-\frac{3}{2}\right)\amp=\multiplyleft{10}\left(-\frac{1}{2}z+\frac{4}{5}\right)\\ 10\left(-\frac{2}{5}z\right)-10\left(\frac{3}{2}\right)\amp=10\left(-\frac{1}{2}z\right)+10\left(\frac{4}{5}\right)\\ -4z-15\amp=-5z+8\\ z-15\amp=8\\ z\amp=23 \end{align*}

Checking the solution $23$ in the equation $-\frac{2}{5}z-\frac{3}{2}=-\frac{1}{2}z+\frac{4}{5}\text{,}$ we replace every instance of $z$ with $23$ and find:

\begin{align*} -\frac{2}{5}z-\frac{3}{2}\amp=-\frac{1}{2}z+\frac{4}{5}\\ -\frac{2}{5}(\substitute{23})-\frac{3}{2}\amp\stackrel{?}{=}-\frac{1}{2}(\substitute{23})+\frac{4}{5}\\ -\frac{46}{5}-\frac{3}{2}\amp\stackrel{?}{=}-\frac{23}{2}+\frac{4}{5}\\ -\frac{46}{5}\cdot\frac{2}{2}-\frac{3}{2}\cdot\frac{5}{5}\amp\stackrel{?}{=}-\frac{23}{2}\cdot\frac{5}{5}+\frac{4}{5}\cdot\frac{2}{2}\\ -\frac{92}{10}-\frac{15}{10}\amp\stackrel{?}{=}-\frac{115}{10}+\frac{8}{10}\\ -\frac{107}{10}\amp\stackrel{\checkmark}{=}-\frac{107}{10} \end{align*}

Thus the solution is $23$ and so the solution set is $\{23\}\text{.}$

Example3.2.5

Solve for $a$ in the equation $\frac{2}{3}(a+1)+5=\frac{1}{3}\text{.}$

Solution

\begin{align*} \frac{2}{3}(a+1)+5\amp=\frac{1}{3}\\ \multiplyleft{3}\left(\frac{2}{3}(a+1)+5\right)\amp=\multiplyleft{3}\frac{1}{3}\\ 3\cdot\frac{2}{3}(a+1)+3\cdot5\amp=1\\ 2(a+1)+15\amp=1\\ 2a+2+15\amp=1\\ 2a+17\amp=1\\ 2a\amp=-16\\ a\amp=-8 \end{align*}

Check the solution $-8$ in the equation $\frac{2}{3}(a+1)+5=\frac{1}{3}\text{,}$ we find that:

\begin{align*} \frac{2}{3}(a+1)+5\amp=\frac{1}{3}\\ \frac{2}{3}(\substitute{-8}+1)+5\amp\stackrel{?}{=}\frac{1}{3}\\ \frac{2}{3}(-7)+5\amp\stackrel{?}{=}\frac{1}{3}\\ -\frac{14}{3}+\frac{15}{3}\amp\stackrel{?}{=}\frac{1}{3}\\ \frac{1}{3}\amp\stackrel{\checkmark}{=}\frac{1}{3} \end{align*}

The solution is therefore $-8$ and the solution set is $\{-8\}\text{.}$

Example3.2.6

Solve for $b$ in the equation $\frac{2b+1}{3}=\frac{2}{5}\text{.}$

Solution

\begin{align*} \frac{2b+1}{3}\amp=\frac{2}{5}\\ \multiplyleft{15}\frac{2b+1}{3}\amp=\multiplyleft{15}\frac{2}{5}\\ 5(2b+1)\amp=6\\ 10b+5\amp=6\\ 10b\amp=1\\ b\amp=\frac{1}{10} \end{align*}

Checking the solution $\frac{1}{10}$ in the equation $\frac{2b+1}{3}=\frac{2}{5}\text{,}$ we find that:

\begin{align*} \frac{2b+1}{3}\amp=\frac{2}{5}\\ \frac{2\left(\substitute{\frac{1}{10}}\right)+1}{3}\amp\stackrel{?}{=}\frac{2}{5}\\ \frac{\frac{1}{5}+1}{3}\amp\stackrel{?}{=}\frac{2}{5}\\ \frac{\frac{1}{5}+\frac{5}{5}}{3}\amp\stackrel{?}{=}\frac{2}{5}\\ \frac{\frac{6}{5}}{3}\amp\stackrel{?}{=}\frac{2}{5}\\ \frac{6}{5}\cdot \frac{1}{3}\amp\stackrel{?}{=}\frac{2}{5}\\ \frac{2}{5}\amp\stackrel{\checkmark}{=}\frac{2}{5} \end{align*}

The solution is $\frac{1}{10}$ and the solution set is $\left\{\frac{1}{10}\right\}\text{.}$

Remark3.2.7

Some of you might solve Example 3.2.6 with a property called cross-multiplication.

Cross-multiplication is a specialized application of the process of clearing the denominators from an equation. This process will be covered in Section 3.4, as there are some restrictions on the process we need to be aware of before making significant use of cross-multiplication.

Let's look at a one more application problem involving equations with fractions.

Example3.2.8

In a science lab, a container had $21$ ounces of water at 9:00AM. Water has been evaporating at the rate of $3$ ounces every $5$ minutes. When will there be $8$ ounces of water left?

Solution

Since the container has been losing $3$ oz of water every $5$ minutes, it loses $\frac{3}{5}$ oz every minute. In $m$ minutes since 9:00AM, the container would lose $\frac{3}{5}m$ oz of water. Since the container had $21$ oz of water at the beginning, the amount of water in the container can be modeled by $21-\frac{3}{5}m$ (in oz).

To find when there would be $8$ oz of water left, we write and solve this equation:

\begin{align*} 21-\frac{3}{5}m\amp=8\\ \multiplyleft{5}(21-\frac{3}{5}x)\amp=\multiplyleft{5}8\\ 5\cdot21-5\cdot\frac{3}{5}x\amp=40\\ 105-3m\amp=40\\ 105-3m\subtractright{105}\amp=40\subtractright{105}\\ -3m\amp=-55\\ \divideunder{-3m}{-3}\amp=\divideunder{-65}{-3}\\ m\amp=\frac{65}{3}\\ m\amp=21\frac{2}{3} \end{align*}

Checking the solution $\frac{65}{3}$ in the equation $21-\frac{3}{5}m=8\text{:}$

\begin{align*} 21-\frac{3}{5}m\amp=8\\ 21-\frac{3}{5}\left(\substitute{\frac{65}{3}}\right)\amp\stackrel{?}{=}8\\ 21-13\amp\stackrel{?}{=}8\\ 8\amp\stackrel{\checkmark}{=}8 \end{align*}

Therefore the solution is $\frac{65}{3}$ or $21\frac{2}{3}\text{.}$ In context, this means that shortly before 9:22AM (or at 9:21:40AM), the container will have $8$ ounces of water left.

Checkpoint3.2.9

Subsection3.2.3Solving Inequalities with Fractions

We can also solve linear inequalities involving fractions by multiplying each side of the inequality by the LCD of all fractions within the inequality. Remember that with inequalities, everything works exactly the same except that the inequality sign reverses direction whenever we multiply each side of the inequality by a negative number.

Example3.2.10

Solve for $x$ in the inequality $\frac{3}{4}x-2\gt\frac{4}{5}x\text{.}$ Write the solution set in both set-builder notation and interval notation.

Solution

\begin{align*} \frac{3}{4}x-2\amp\gt\frac{4}{5}x\\ \multiplyleft{20}\left(\frac{3}{4}x-2\right)\amp\gt\multiplyleft{20}\frac{4}{5}x\\ 20\cdot\frac{3}{4}x-20\cdot2\amp\gt16x\\ 15x-40\amp\gt16x\\ 15x-40\subtractright{15x}\amp\gt16x\subtractright{15x}\\ -40\amp\gt x\\ x\amp\lt-40 \end{align*}

The solution set in set-builder notation is $\{x\mid x\lt-40\}\text{.}$ Note that it's equivalent to write $\{x\mid-40\gt x\}\text{,}$ but it's easier to understand if we write $x$ first in an inequality.

The solution set in interval notation is $(-\infty,-40)\text{.}$

Example3.2.11

Solve for $y$ in the inequality $\frac{4}{7}-\frac{4}{3}y\le\frac{2}{3}\text{.}$ Write the solution set in both set-builder notation and interval notation.

Solution

\begin{align*} \frac{4}{7}-\frac{4}{3}y\amp\le\frac{2}{3}\\ \multiplyleft{21}\left(\frac{4}{7}-\frac{4}{3}y\right)\amp\le\multiplyleft{21}\left(\frac{2}{3}\right)\\ 21\left(\frac{4}{7}\right)-21\left(\frac{4}{3}y\right)\amp\le21\left(\frac{2}{3}\right)\\ 12-28y\amp\le14\\ -28y\amp\le2\\ \divideunder{-28y}{-28}\amp\ge\divideunder{2}{-28}\\ y\amp\ge-\frac{1}{14} \end{align*}

Note that when we divided each side of the inequality by $-28\text{,}$ the inequality symbol reversed direction.

The solution set in set-builder notation is $\left\{y\mid y\ge-\frac{1}{14}\right\}\text{.}$

The solution set in interval notation is $\left[-\frac{1}{14},\infty\right)\text{.}$

Example3.2.12

In a certain class, a student's grade is calculated by the average of their scores on 3 tests. Ronda scored $78\%$ and $54\%$ on the first two tests. If Ronda wants to earn at least a grade of C $(70\%)\text{,}$ what's the lowest score she needs to earn on the third exam?

Solution

Assume Ronda will score $x\%$ on the third test. To make her average test score greater than or equal to $70\%\text{,}$ we write and solve this inequality:

\begin{align*} \frac{78+54+x}{3}\amp\ge70\\ \frac{132+x}{3}\amp\ge70\\ \multiplyleft{3}\frac{132+x}{3}\amp\ge\multiplyleft{3}70\\ 132+x\amp\ge210\\ x\amp\ge78 \end{align*}

To earn at least a C grade, Ronda needs to score at least $78\%$ on the third test.

SubsectionExercises

Solving Linear Equations with Fractions Exercises

1

Solve the equation.

$\displaystyle{ {\frac{C}{5}+45}={2C} }$

2

Solve the equation.

$\displaystyle{ {\frac{n}{2}+18}={5n} }$

3

Solve the equation.

$\displaystyle{ {\frac{p}{8}+6}={10} }$

4

Solve the equation.

$\displaystyle{ {\frac{x}{5}+3}={9} }$

5

Solve the equation.

$\displaystyle{ {-\frac{y}{9}+7}={5} }$
$\displaystyle{ {\frac{-x}{9}+7}={5} }$
$\displaystyle{ {\frac{a}{-9}+7}={5} }$
$\displaystyle{ {\frac{-p}{-9}+7}={5} }$

6

Solve the equation.

$\displaystyle{ {-\frac{t}{5}+1}={-1} }$
$\displaystyle{ {\frac{-C}{5}+1}={-1} }$
$\displaystyle{ {\frac{m}{-5}+1}={-1} }$
$\displaystyle{ {\frac{-p}{-5}+1}={-1} }$

7

Solve the equation.

$\displaystyle{ {4-\frac{b}{3}} = {0} }$

8

Solve the equation.

$\displaystyle{ {10-\frac{c}{7}} = {8} }$

9

Solve the equation.

$\displaystyle{ {-53} = {7-\frac{10B}{3}} }$

10

Solve the equation.

$\displaystyle{ {-45} = {3-\frac{8C}{5}} }$

11

Solve the equation.

$\displaystyle{ {5n} = {\frac{9n}{10}+123} }$

12

Solve the equation.

$\displaystyle{ {5p} = {\frac{5p}{6}+50} }$

13

Solve the equation.

$\displaystyle{ {155} = {{\frac{10}{7}}x+3x} }$

14

Solve the equation.

$\displaystyle{ {78} = {{\frac{8}{3}}y+6y} }$

15

Solve the equation.

$\displaystyle{ {95-{\frac{7}{6}}t} = {2t} }$

16

Solve the equation.

$\displaystyle{ {145-{\frac{9}{4}}b} = {5b} }$

17

Solve the equation.

$\displaystyle{ {6c} = {{\frac{8}{5}}c+9} }$

18

Solve the equation.

$\displaystyle{ {3B} = {{\frac{8}{9}}B+6} }$

19

Solve the equation.

$\displaystyle{ {{\frac{3}{2}}-3C}={8} }$

20

Solve the equation.

$\displaystyle{ {{\frac{7}{10}}-9n}={8} }$

21

Solve the equation.

$\displaystyle{ {{\frac{7}{6}}-{\frac{1}{6}}p}={7} }$

22

Solve the equation.

$\displaystyle{ {{\frac{5}{2}}-{\frac{1}{2}}x}={4} }$

23

Solve the equation.

$\displaystyle{ {\frac{8y}{9}-3}={-{\frac{107}{9}}} }$

24

Solve the equation.

$\displaystyle{ {\frac{4t}{5}-9}={-{\frac{33}{5}}} }$

25

Solve the equation.

$\displaystyle{ {{\frac{8}{5}}+{\frac{6}{5}}a}={7a} }$

26

Solve the equation.

$\displaystyle{ {{\frac{2}{9}}+{\frac{4}{9}}c}={7c} }$

27

Solve the equation.

$\displaystyle{ {\frac{3B}{7}-{\frac{32}{7}}}={-{\frac{5}{7}}B} }$

28

Solve the equation.

$\displaystyle{ {\frac{2C}{3}-{\frac{28}{3}}}={-{\frac{5}{3}}C} }$

29

Solve the equation.

$\displaystyle{ {\frac{4n}{7}+{\frac{5}{4}}}={n} }$

30

Solve the equation.

$\displaystyle{ {\frac{4p}{5}+{\frac{5}{2}}}={p} }$

31

Solve the equation.

$\displaystyle{ {\frac{2x}{7}-94}={-{\frac{5}{6}}x} }$

32

Solve the equation.

$\displaystyle{ {\frac{4y}{5}-31}={-{\frac{3}{4}}y} }$

33

Solve the equation.

$\displaystyle{ {-{\frac{7}{4}}t+17}={\frac{3t}{8}} }$

34

Solve the equation.

$\displaystyle{ {-{\frac{7}{4}}a+34}={\frac{3a}{8}} }$

35

Solve the equation.

$\displaystyle{ {\frac{7c}{6}-9c}={{\frac{5}{12}}} }$

36

Solve the equation.

$\displaystyle{ {\frac{5B}{4}-7B}={{\frac{5}{8}}} }$

37

Solve the equation.

$\displaystyle{ {\frac{7C}{6}+{\frac{8}{5}}}={{\frac{5}{8}}C} }$

38

Solve the equation.

$\displaystyle{ {\frac{5n}{4}+{\frac{4}{9}}}={{\frac{5}{6}}n} }$

39

Solve the equation.

$\displaystyle{ {{\frac{4}{3}}p}={{\frac{1}{5}}+\frac{3p}{4}} }$

40

Solve the equation.

$\displaystyle{ {{\frac{2}{7}}x}={{\frac{1}{3}}+\frac{5x}{4}} }$

41

Solve the equation.

$\displaystyle{ {{\frac{5}{6}}}={\frac{y}{12}} }$

42

Solve the equation.

$\displaystyle{ {{\frac{9}{2}}}={\frac{t}{10}} }$

43

Solve the equation.

$\displaystyle{ {-\frac{a}{36}}={{\frac{2}{9}}} }$

44

Solve the equation.

$\displaystyle{ {-\frac{c}{15}}={{\frac{8}{5}}} }$

45

Solve the equation.

$\displaystyle{ {-\frac{A}{12}}={-{\frac{3}{2}}} }$

46

Solve the equation.

$\displaystyle{ {-\frac{C}{40}}={-{\frac{5}{8}}} }$

47

Solve the equation.

$\displaystyle{ {-{\frac{8}{5}}}={\frac{7n}{4}} }$

48

Solve the equation.

$\displaystyle{ {-{\frac{3}{2}}}={\frac{9p}{4}} }$

49

Solve the equation.

$\displaystyle{ {{\frac{7}{8}}}={\frac{x+9}{24}} }$

50

Solve the equation.

$\displaystyle{ {{\frac{3}{4}}}={\frac{y+6}{12}} }$

51

Solve the equation.

$\displaystyle{ \frac{7}{2} = \frac{t-3}{5} }$

52

Solve the equation.

$\displaystyle{ \frac{9}{8} = \frac{a-3}{9} }$

53

Solve the equation.

$\displaystyle{ {\frac{c-4}{4}}={\frac{c+8}{6}} }$

54

Solve the equation.

$\displaystyle{ {\frac{A-8}{6}}={\frac{A+5}{8}} }$

55

Solve the equation.

$\displaystyle{ {\frac{C+2}{4}-\frac{C-2}{8}}={{\frac{5}{4}}} }$

56

Solve the equation.

$\displaystyle{ {\frac{n+7}{2}-\frac{n-9}{4}}={{\frac{17}{4}}} }$

57

Solve the equation.

$\displaystyle{ {\frac{7p+6}{4}-\frac{4-p}{8}}={{\frac{2}{3}}} }$

58

Solve the equation.

$\displaystyle{ {\frac{5x+6}{4}-\frac{1-x}{8}}={{\frac{4}{5}}} }$

59

Solve the equation.

$\displaystyle{ {3}={\frac{y}{5}+\frac{y}{10}} }$

60

Solve the equation.

$\displaystyle{ {39}={\frac{t}{9}+\frac{t}{4}} }$

61

Solve the equation.

$\displaystyle{ {\frac{a}{5}-8}={\frac{a}{9}} }$

62

Solve the equation.

$\displaystyle{ {\frac{c}{3}-2}={\frac{c}{5}} }$

63

Solve the equation.

$\displaystyle{ {\frac{A}{6}-2}={\frac{A}{9}+1} }$

64

Solve the equation.

$\displaystyle{ {\frac{C}{4}-2}={\frac{C}{10}+7} }$

65

Solve the equation.

$\displaystyle{ {{\frac{3}{8}}m+{\frac{5}{8}}}={{\frac{5}{8}}m+{\frac{1}{4}}} }$

66

Solve the equation.

$\displaystyle{ {{\frac{4}{5}}p+1}={{\frac{7}{5}}p+{\frac{7}{5}}} }$

67

Solve the equation.

$\displaystyle{ {-{\frac{3}{4}}x-{\frac{9}{2}}}={{\frac{8}{7}}x-{\frac{3}{2}}} }$

68

Solve the equation.

$\displaystyle{ {y-{\frac{7}{8}}}={{\frac{3}{7}}y+{\frac{3}{5}}} }$

69

Solve the equation.

$\displaystyle{ {-{\frac{8}{7}}t-1}={t+{\frac{7}{4}}} }$

70

Solve the equation.

$\displaystyle{ {-{\frac{5}{2}}a-{\frac{1}{5}}}={{\frac{3}{7}}a-{\frac{2}{3}}} }$

Application Problems

71

Marc is jogging in a straight line. He got a head start of $9$ meters from the starting line, and he ran $3$ meters every $4$ seconds. After how many seconds will Marc be $21$ meters away from the starting line?

Marc will be $21$ meters away from the starting line seconds since she started running.

72

Chris is jogging in a straight line. He started at a place $35$ meters from the starting line, and ran toward the starting line at the speed of $4$ meters every $9$ seconds. After how many seconds will Chris be $23$ meters away from the starting line?

Chris will be $23$ meters away from the starting line seconds since she started running.

73

Morah had only ${\$9.00}$ in her piggy bank, and she decided to start saving more. She saves ${\$4.00}$ every $7$ days. After how many days will she have ${\$17.00}$ in the piggy bank?

Morah will save ${\$17.00}$ in her piggy bank after days.

74

Carmen has saved ${\$41.00}$ in her piggy bank, and she decided to start spending them. She spends ${\$5.00}$ every $6$ days. After how many days will she have ${\$16.00}$ left in the piggy bank?

Carmen will have ${\$16.00}$ left in her piggy bank after days.

Solving Inequalities with Fractions Exercises