ORCCA Linear Equations and Inequalities with Fractions

Section 3.3 Linear Equations and Inequalities with Fractions

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

Subsection 3.3.1 Introduction

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.

Subsection 3.3.2 Eliminating Denominators

Example 3.3.1.

Deshawn planted a sapling in his yard that was $4$-feet tall. The tree will grow $\frac{2}{3}$ of a foot every year. How many years will it take for his tree to be $10$ feet tall?

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 Deshawn's 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 \end{align*}

In summary, it will take $9$ years for Deshawn's tree to reach $10$ feet tall.

Let's look at a few more examples.

Example 3.3.2.

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

Explanation

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 $-2\text{:}$

\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 $\{-2\}\text{.}$

Example 3.3.3.

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

Explanation

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 $23\text{:}$

\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 $\{23\}\text{.}$

Example 3.3.4.

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

Explanation

\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{\checkmark}{=}\frac{1}{3} \end{align*}

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

Example 3.3.5.

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

Explanation

\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 $\frac{1}{10}\text{:}$

\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{\checkmark}{=}\frac{2}{5} \end{align*}

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

Remark 3.3.6.

You might know about solving Example 3.3.5 with a technique called cross-multiplication. Cross-multiplication is a specialized application of the process of clearing the denominators from an equation. This process will be discussed in Section 3.5.

Example 3.3.7.

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

Explanation

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 9:00 AM, 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}\left(21-\frac{3}{5}x\right)\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=-65\\ \divideunder{-3m}{-3}\amp=\divideunder{-65}{-3}\\ m\amp=\frac{65}{3} \end{align*}

Checking the solution $\frac{65}{3}\text{:}$

\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{\checkmark}{=}8 \end{align*}

Therefore, the solution is $\frac{65}{3}\text{.}$ As a mixed number, this is $21\frac{2}{3}\text{.}$ In context, this means that $21$ minutes and $40$ seconds after 9:00 AM, at 9:21:40 AM, the container will have $8$ ounces of water left.

Checkpoint 3.3.8.

Subsection 3.3.3 Solving 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 linear inequalities, everything works exactly the same as with solving linear equations, except that the inequality sign reverses direction whenever we multiply each side of the inequality by a negative number.

Let's look at a few examples to see how to solve a linear inequality by first clearing the fractions.

Example 3.3.9.

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.

Explanation

\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 $(-\infty,-40)\text{.}$

Example 3.3.10.

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.

Explanation

\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 $\left[-\frac{1}{14},\infty\right)\text{.}$

Example 3.3.11.

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

Explanation

Assume Aidan will score $x\%$ on the third test. To make his 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, Aidan needs to score at least $78\%$ on the third test.

Exercises 3.3.4 Exercises

Review and Warmup

1.

Multiply: $\displaystyle{8\cdot \frac{2}{3} }$

2.

Multiply: $\displaystyle{3\cdot \frac{3}{8} }$

3.

Multiply: $\displaystyle{4\cdot\left( -{\frac{5}{2}} \right)}$

4.

Multiply: $\displaystyle{28\cdot\left( -{\frac{6}{7}} \right)}$

5.

Do the following multiplications.

$14 \cdot \frac{5}{7}$
$21 \cdot \frac{5}{7}$
$28 \cdot \frac{5}{7}$

6.

Do the following multiplications.

$27 \cdot \frac{5}{9}$
$36 \cdot \frac{5}{9}$
$45 \cdot \frac{5}{9}$

Solving Linear Equations with Fractions

7.

Solve the equation.

$\displaystyle{ {\frac{q}{7}+120}={3q} }$

8.

Solve the equation.

$\displaystyle{ {\frac{y}{4}+57}={5y} }$

9.

Solve the equation.

$\displaystyle{ {\frac{t}{10}+8}={13} }$

10.

Solve the equation.

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

11.

Solve the equation.

$\displaystyle{ {3-\frac{c}{7}} = {1} }$

12.

Solve the equation.

$\displaystyle{ {9-\frac{A}{2}} = {3} }$

13.

Solve the equation.

$\displaystyle{ {-42} = {6-\frac{8C}{5}} }$

14.

Solve the equation.

$\displaystyle{ {-22} = {2-\frac{4m}{9}} }$

15.

Solve the equation.

$\displaystyle{ {5p} = {\frac{3p}{10}+282} }$

16.

Solve the equation.

$\displaystyle{ {5q} = {\frac{7q}{6}+115} }$

17.

Solve the equation.

$\displaystyle{ {64} = {{\frac{10}{3}}y+2y} }$

18.

Solve the equation.

$\displaystyle{ {195} = {{\frac{4}{7}}r+5r} }$

19.

Solve the equation.

$\displaystyle{ {74-{\frac{7}{6}}a} = {5a} }$

20.

Solve the equation.

$\displaystyle{ {65-{\frac{5}{2}}c} = {4c} }$

21.

Solve the equation.

$\displaystyle{ {5A} = {{\frac{4}{9}}A+8} }$

22.

Solve the equation.

$\displaystyle{ {3C} = {{\frac{2}{3}}C+4} }$

23.

Solve the equation.

$\displaystyle{ {{\frac{7}{2}}-9m}={8} }$

24.

Solve the equation.

$\displaystyle{ {{\frac{3}{8}}-7p}={8} }$

25.

Solve the equation.

$\displaystyle{ {{\frac{5}{4}}-{\frac{1}{4}}q}={3} }$

26.

Solve the equation.

$\displaystyle{ {{\frac{9}{2}}-{\frac{1}{2}}y}={1} }$

27.

Solve the equation.

$\displaystyle{ {\frac{4r}{7}-8}={-{\frac{40}{7}}} }$

28.

Solve the equation.

$\displaystyle{ {\frac{6a}{5}-6}={-{\frac{6}{5}}} }$

29.

Solve the equation.

$\displaystyle{ {{\frac{2}{9}}+{\frac{8}{9}}c}={7c} }$

30.

Solve the equation.

$\displaystyle{ {{\frac{6}{7}}+{\frac{10}{7}}A}={7A} }$

31.

Solve the equation.

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

32.

Solve the equation.

$\displaystyle{ {\frac{5m}{11}-{\frac{21}{11}}}={-{\frac{2}{11}}m} }$

33.

Solve the equation.

$\displaystyle{ {\frac{6p}{7}+{\frac{7}{2}}}={p} }$

34.

Solve the equation.

$\displaystyle{ {\frac{2q}{3}+{\frac{5}{8}}}={q} }$

35.

Solve the equation.

$\displaystyle{ {\frac{4y}{7}-111}={-{\frac{3}{4}}y} }$

36.

Solve the equation.

$\displaystyle{ {\frac{2r}{5}-19}={-{\frac{3}{2}}r} }$

37.

Solve the equation.

$\displaystyle{ {-{\frac{5}{8}}a+39}={\frac{3a}{16}} }$

38.

Solve the equation.

$\displaystyle{ {-{\frac{7}{10}}b+51}={\frac{3b}{20}} }$

39.

Solve the equation.

$\displaystyle{ {\frac{5A}{6}-8A}={{\frac{5}{12}}} }$

40.

Solve the equation.

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

41.

Solve the equation.

$\displaystyle{ {\frac{7m}{6}+{\frac{8}{9}}}={{\frac{3}{4}}m} }$

42.

Solve the equation.

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

43.

Solve the equation.

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

44.

Solve the equation.

$\displaystyle{ {{\frac{5}{6}}y}={{\frac{5}{7}}+\frac{3y}{5}} }$

45.

Solve the equation.

$\displaystyle{ {{\frac{7}{6}}}={\frac{r}{24}} }$

46.

Solve the equation.

$\displaystyle{ {{\frac{3}{2}}}={\frac{a}{8}} }$

47.

Solve the equation.

$\displaystyle{ {-\frac{b}{14}}={{\frac{8}{7}}} }$

48.

Solve the equation.

$\displaystyle{ {-\frac{A}{30}}={{\frac{2}{5}}} }$

49.

Solve the equation.

$\displaystyle{ {-\frac{C}{10}}={-{\frac{5}{2}}} }$

50.

Solve the equation.

$\displaystyle{ {-\frac{m}{24}}={-{\frac{9}{8}}} }$

51.

Solve the equation.

$\displaystyle{ {-{\frac{3}{4}}}={\frac{7p}{5}} }$

52.

Solve the equation.

$\displaystyle{ {-{\frac{7}{10}}}={\frac{5q}{9}} }$

53.

Solve the equation.

$\displaystyle{ {{\frac{3}{8}}}={\frac{y+6}{24}} }$

54.

Solve the equation.

$\displaystyle{ {{\frac{7}{4}}}={\frac{r+3}{12}} }$

55.

Solve the equation.

$\displaystyle{ \frac{9}{10} = \frac{a-4}{9} }$

56.

Solve the equation.

$\displaystyle{ \frac{5}{6} = \frac{b-4}{7} }$

57.

Solve the equation.

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

58.

Solve the equation.

$\displaystyle{ {\frac{B-2}{6}}={\frac{B+2}{8}} }$

59.

Solve the equation.

$\displaystyle{ {\frac{m+7}{4}-\frac{m-9}{8}}={{\frac{9}{4}}} }$

60.

Solve the equation.

$\displaystyle{ {\frac{p+1}{2}-\frac{p-6}{4}}={{\frac{3}{4}}} }$

61.

Solve the equation.

$\displaystyle{ {\frac{q}{7}-2}={\frac{q}{9}} }$

62.

Solve the equation.

$\displaystyle{ {\frac{y}{5}-1}={\frac{y}{10}} }$

63.

Solve the equation.

$\displaystyle{ {\frac{r}{2}-3}={\frac{r}{5}+6} }$

64.

Solve the equation.

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

65.

Solve the equation.

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

66.

Solve the equation.

$\displaystyle{ {1A+3}={5A+{\frac{3}{2}}} }$

67.

Solve the equation.

$\displaystyle{ {\frac{7B+6}{4}-\frac{4-B}{8}}={{\frac{1}{9}}} }$

68.

Solve the equation.

$\displaystyle{ {\frac{5m+6}{2}-\frac{1-m}{4}}={{\frac{1}{5}}} }$

69.

Solve the equation.

$\displaystyle{ {11}={\frac{p}{3}+\frac{p}{8}} }$

70.

Solve the equation.

$\displaystyle{ {27}={\frac{q}{7}+\frac{q}{2}} }$

71.

Solve the equation.

$\displaystyle{ {{\frac{5}{9}}y-{\frac{6}{7}}}={-{\frac{7}{2}}y+{\frac{1}{2}}} }$

72.

Solve the equation.

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

73.

Solve the equation.

$\displaystyle{ {-{\frac{7}{8}}a-{\frac{10}{9}}}={{\frac{7}{2}}a-{\frac{5}{2}}} }$

74.

Solve the equation.

$\displaystyle{ {-{\frac{5}{9}}b+{\frac{5}{7}}}={{\frac{5}{3}}b-1} }$

75.

Solve the equation.

$\displaystyle{ {-\frac{A}{2}+8}={4} }$
$\displaystyle{ {\frac{-q}{2}+8}={4} }$
$\displaystyle{ {\frac{t}{-2}+8}={4} }$
$\displaystyle \displaystyle{ {\frac{-c}{-2}+8}={4} }$

76.

Solve the equation.

$\displaystyle{ {-\frac{B}{2}+2}={-4} }$
$\displaystyle{ {\frac{-C}{2}+2}={-4} }$
$\displaystyle{ {\frac{r}{-2}+2}={-4} }$
$\displaystyle \displaystyle{ {\frac{-x}{-2}+2}={-4} }$

Applications

77.

Lindsay is jogging in a straight line. She got a head start of $10$ meters from the starting line, and she ran $2$ meters every $7$ seconds. After how many seconds will Lindsay be $18$ meters away from the starting line?

Lindsay will be $18$ meters away from the starting line seconds since she started running.

78.

Amber is jogging in a straight line. She started at a place $30$ meters from the starting line, and ran toward the starting line at the speed of $2$ meters every $3$ seconds. After how many seconds will Amber be $26$ meters away from the starting line?

Amber will be $26$ meters away from the starting line seconds since she started running.

79.

Joseph had only ${\$10.00}$ in his piggy bank, and he decided to start saving more. He saves ${\$5.00}$ every $8$ days. After how many days will he have ${\$25.00}$ in the piggy bank?

Joseph will save ${\$25.00}$ in his piggy bank after days.

80.

Ross has saved ${\$50.00}$ in his piggy bank, and he decided to start spending them. He spends ${\$3.00}$ every $4$ days. After how many days will he have ${\$44.00}$ left in the piggy bank?

Ross will have ${\$44.00}$ left in his piggy bank after days.

Solving Inequalities with Fractions

For each problem below, solve the inequality and, in addition to writing both the set-builder and interval notations for each solution set, draw a graph of each solution set on a number line.

81.

${\frac{x}{3}+84} \geq {5x}$

;

82.

${\frac{x}{4}+28} \geq {2x}$

;

83.

${{\frac{3}{4}}-5y} \lt {3}$

;

84.

${{\frac{5}{4}}-2y} \lt {6}$

;

85.

${-{\frac{3}{4}}t} > {{\frac{4}{5}}t-93}$

;

86.

${-{\frac{5}{6}}t} > {{\frac{2}{7}}t-47}$

;

87.

;

88.

;

89.

${-\frac{z}{4}} \lt {-\frac{3}{2}}$

;

90.

${-\frac{z}{8}} \lt {-\frac{9}{2}}$

;

91.

${\frac{x}{7}-12} \leq {\frac{x}{3}}$

;

92.

${\frac{x}{7}-4} \leq {\frac{x}{5}}$

;

93.

${\frac{y-8}{6}} \geq {\frac{y+8}{4}}$

;

94.

${\frac{y-4}{6}} \geq {\frac{y+3}{4}}$

;

95.

${{\frac{17}{12}}} \lt {\frac{x+1}{6}-\frac{x-7}{12}}$

;

96.

${{\frac{17}{12}}} \lt {\frac{x+7}{6}-\frac{x-1}{12}}$

;

Applications

97.

Your grade in a class is determined by the average of three test scores. You scored $75$ and $86$ on the first two tests. To earn at least $80$ for this course, how much do you have to score on the third test? Let $x$ be the score you will earn on the third test.

Write an inequality to represent this situation.
Solve this inequality. What is the minimum that you have to earn on the third test in order to earn a $80$ for the course?
You cannot score over $100$ on the third test. Use interval notation to represent the range of scores you can earn on the third test in order to earn at least $80$ for this course.

98.

Your grade in a class is determined by the average of three test scores. You scored $70$ and $90$ on the first two tests. To earn at least $83$ for this course, how much do you have to score on the third test? Let $x$ be the score you will earn on the third test.

Write an inequality to represent this situation.
Solve this inequality. What is the minimum that you have to earn on the third test in order to earn a $83$ for the course?
You cannot score over $100$ on the third test. Use interval notation to represent the range of scores you can earn on the third test in order to earn at least $83$ for this course.

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