ORCCA Linear Equations and Inequalities with Fractions

Section 2.3 Linear Equations and Inequalities with Fractions

Objectives: PCC Course Content and Outcome Guide

permalinkThis section discusses a technique that might make it easier for you to solve linear equations and inequalities when there are fractions present.

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Figure 2.3.1. Alternative Video Lessons

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Subsection 2.3.1 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{aligned} 2 x & = 10 \\ \frac{2 x}{2} & = \frac{10}{2} \\ x & = 5 \end{aligned}

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

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

\begin{aligned} \frac{1}{2} x & = 10 \\ \frac{\frac{1}{2} x}{\frac{1}{2}} & = \frac{10}{\frac{1}{2}} \\ x & = 10 \cdot \frac{2}{1} = 20 \end{aligned}

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

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 just the right number that eliminates the denominator(s). 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{aligned} \frac{1}{2} x & = 10 \\ 2 \cdot (\frac{1}{2} x) & = 2 \cdot 10 \\ x & = 20 \end{aligned}

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

permalinkFor more complicated equations, multiply each side of the equation by the least common denominator (LCD) of all fractions appearing in the equation. In this way we can take one step to turn an equation full of fractions into an equation with no fractions.

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Subsection 2.3.2 Eliminating Denominators

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Example 2.3.2.

Deshawn planted a sapling in his yard that was 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 ft 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 set up the equation

4 + \frac{2}{3} y = 10

$\begin{equation*} 4+\frac{2}{3}y=10 \end{equation*}$

To solve the equation, we take note of the fraction and its denominator $3\text{.}$ As the very first step, multiplying by $3$ on each side will leave us with no fractions.

\begin{aligned} 4 + \frac{2}{3} y & = 10 \\ 3 \cdot (4 + \frac{2}{3} y) & = 3 \cdot 10 \\ 3 \cdot 4 + 3 \cdot \frac{2}{3} y & = 30 \\ 12 + 2 y & = 30 \\ 2 y & = 18 \\ y & = 9 \end{aligned}

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

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

\begin{aligned} 4 + \frac{2}{3} y & = 10 \\ 4 + \frac{2}{3} (9) & \overset{?}{=} 10 \\ 4 + 6 & \overset{✓}{=} 10 \end{aligned}

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

In summary, it will take $9$ years for Deshawn's tree to reach $10$ feet tall. The point of this example was to demonstrate that clearing denominators can make an equation fairly easy to solve.

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Example 2.3.3.

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

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Checkpoint 2.3.4.

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Example 2.3.5.

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

Explanation

There are two fractions appearing here, but both have the same denominator $3\text{.}$ So we can multiply by $3$ on each side to clear denominators.

\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 $\{-8\}\text{.}$

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Example 2.3.6.

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

Explanation

The structure of the equation is a little different from previous examples, but we can still see two denominators $3$ and $5\text{,}$ and find their LCM is $15\text{.}$

\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 $\left\{\frac{1}{10}\right\}\text{.}$

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Example 2.3.7.

Suppose we want to know the total cost for a box of cereal that weighs $18$ ounces, assuming it costs the same per ounce as the $21$ -ounce box. Letting $C$ be this unknown cost (in dollars), we could set up the following proportion:

\begin{aligned} \frac{cost in dollars}{weight in oz} & = \frac{cost in dollars}{weight in oz} \\ \frac{$ 3.99}{21 oz} & = \frac{$ C}{18 oz} \end{aligned}

$\begin{align*} \frac{\text{cost in dollars}}{\text{weight in oz}}\amp=\frac{\text{cost in dollars}}{\text{weight in oz}}\\ \frac{\$3.99}{21\,\text{oz}}\amp=\frac{\$C}{18\,\text{oz}} \end{align*}$

We take a moment to rewrite the equation without units:

\frac{3.99}{21} = \frac{C}{18}

$\begin{equation*} \frac{3.99}{21}=\frac{C}{18} \end{equation*}$

Next we want to recognize that each side contains a fraction. Our usual approach for solving this type of equation is to multiply each side by the least common denominator (LCD). In this case, the LCD of $21$ and $18$ is $126\text{.}$ As with many other proportions we solve, it is often easier to just multiply each side by the common denominator of $18\cdot 21\text{,}$ which we know will make each denominator cancel:

\begin{aligned} \frac{3.99}{21} & = \frac{C}{18} \\ 18 \cdot 21 \cdot \frac{3.99}{21} & = \frac{C}{18} \cdot 18 \cdot 21 \\ 18 \cdot 21 \frac{3.99}{21} & = \frac{C}{18} \cdot 18 \cdot 21 \\ 71.82 & = 21 C \\ \frac{71.82}{21} & = \frac{21 C}{21} \\ C & = 3.42 \end{aligned}

$\begin{align*} \frac{3.99}{21}\amp=\frac{C}{18}\\ \multiplyleft{18\cdot21}\frac{3.99}{21}\amp=\frac{C}{18}\multiplyright{18\cdot21}\\ 18\cdot\cancelhighlight{21}\frac{3.99}{\cancelhighlight{21}}\amp=\frac{C}{\cancelhighlight{18}}\cdot\cancelhighlight{18}\cdot21\\ 71.82\amp=21C\\ \divideunder{71.82}{21}\amp=\divideunder{21C}{21}\\ C\amp=3.42 \end{align*}$

So assuming the cost is proportional to the cost of the $21$ -ounce box, the cost for an $18$ -ounce box of cereal would be $\$3.42\text{.}$

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Example 2.3.8.

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.

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Subsection 2.3.3 Creating and Solving Proportions

permalinkProportions can be used to solve many real-life applications where two quantities vary together. For example, if your home is worth more, your property tax will be more. If you have a larger amount of liquid Tylenol, you have more milligrams of the drug dissolved in that liquid. The key to using proportions is to first set up a ratio where all values are known. We then set up a second ratio that will be proportional to the first, but has an unknown value.

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Example 2.3.9.

Property taxes for a residential property are proportional to the assessed value of the property. A certain home is assessed at $\$234{,}100$ and its annual property taxes are $\$2{,}518.92\text{.}$ What are the annual property taxes for the house next door that is assessed at $\$287{,}500\text{?}$

Explanation

Let $T$ be the annual property taxes (in dollars) for a property assessed at $\$287{,}500\text{.}$ We can write and solve this proportion:

\begin{align*} \frac{\text{tax}}{\text{property value}}\amp=\frac{\text{tax}}{\text{property value}}\\ \frac{2518.92}{234100}\amp=\frac{T}{287500} \end{align*}

The least common denominator of this proportion is rather large, so we will instead multiply each side by $234100$ and $287500$ and simplify from there:

\begin{align*} \frac{2518.92}{234100}\amp=\frac{T}{287500}\\ \multiplyleft{234100\cdot287500}\frac{2518.92}{234100}\amp=\frac{T}{287500}\multiplyright{234100\cdot287500}\\ 287500\cdot2518.92 \amp=T\cdot 234100\\ \frac{287500\cdot 2518.92}{234100} \amp=\frac{234100T}{234100}\\ T\amp\approx3093.50 \end{align*}

The property taxes for a property assessed at $\$287{,}500$ are $\$3{,}093.50\text{.}$

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Example 2.3.10.

Tagging fish is a means of estimating the size of the population of fish in a lake. A sample of fish is taken, tagged, and then redistributed into the lake. Later when another sample is taken, some of those fish will have tags. The number of tagged fish are assumed to be proportional to the total number of fish. We can look at that relationship from the perspective of the entire lake, or just the second sample, and we get two ratios that should be proportional.

\frac{number of tagged fish in sample}{number of fish in sample} = \frac{number of tagged fish total}{number of fish total}

$\begin{equation*} \frac{\text{number of tagged fish in sample}}{\text{number of fish in sample}}=\frac{\text{number of tagged fish total}}{\text{number of fish total}} \end{equation*}$

Assume that $90$ fish are caught and tagged. Once they are redistributed, a sample of $200$ fish is taken. Of these, $7$ are tagged. Estimate how many fish total are in the lake.

Explanation

Let $n$ be the number of fish in the lake. We can set up a proportion for this scenario:

\begin{align*} \frac{7}{200}\amp=\frac{90}{n} \end{align*}

To solve for $n\text{,}$ which is in a denominator, we'll need to multiply each side by both $200$ and $n\text{:}$

\begin{align*} \frac{7}{200}\amp=\frac{90}{n}\\ \multiplyleft{200\cdot n}\frac{7}{200}\amp=\frac{90}{n}\multiplyright{200\cdot n}\\ {\cancelhighlight{200}\cdot n}\cdot\frac{7}{\cancelhighlight{200}}\amp=\frac{90}{\cancelhighlight{n}}\cdot200\cdot\cancelhighlight{n}\\ 7n\amp=1800\\ \frac{7n}{7}\amp=\frac{1800}{7}\\ n\amp\approx 2471.4286 \end{align*}

According to this sample, we can estimate that there are about $2471$ fish in the lake.

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Checkpoint 2.3.11.

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Example 2.3.12.

Sarah is an architect and she's making a scale model of a building. The actual building will be 30 ft tall. In the model, the height of the building will be 2 in. How tall should she make the model of a person who is 5 ft 6 in tall so that the model is to scale?

Explanation

Let $h$ be the height of the person in Sarah's model, which we'll measure in inches. We'll create a proportion that compares the building and person's heights in the model to their heights in real life:

\begin{align*} \frac{\text{height of model building in inches}}{\text{height of actual building in feet}}\amp=\frac{\text{height of model person in inches}}{\text{height of actual person in feet}}\\ \frac{2\,\text{in}}{30\,\text{ft}}\amp=\frac{h\,\text{in}}{5\,\text{ft}\,6\text{in}} \end{align*}

Before we can just eliminate the units, we'll need to convert 5 ft 6 in to feet:

\begin{equation*} \frac{2\,\text{in}}{30\,\text{ft}}=\frac{h\,\text{in}}{5.5\,\text{ft}} \end{equation*}

Now we can remove the units and continue solving:

\begin{align*} \frac{2}{30}\amp=\frac{h}{5.5}\\ \multiplyleft{30\cdot5.5}\frac{2}{30}\amp=\frac{h}{5.5}\multiplyright{30\cdot5.5}\\ 5.5\cdot 2 \amp= h\cdot 30\\ 11\amp=30h\\ \frac{11}{30}\amp=\frac{30h}{30}\\ \frac{11}{30}\amp=h\\ h\amp\approx 0.3667 \end{align*}

Sarah should make the model of a person who is 5 ft 6 in tall be $\frac{11}{30}$ inches (about $0.3667$ inches) tall.

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Subsection 2.3.4 Solving Inequalities with Fractions

permalinkThe notion of clearing denominators can also apply when solving a linear inequality. Remember that with inequalities, the only difference in the process is that the inequality sign reverses direction whenever we multiply or divide each side by a negative number.

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Example 2.3.13.

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

The LCM of the denominators is $20\text{,}$ so we start out multiplying each side of the inequality by $20\text{.}$

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

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Checkpoint 2.3.14.

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Example 2.3.15.

In a certain class, a student's grade is calculated by the average of their scores on three 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.

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Reading Questions 2.3.5 Reading Questions

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1.

What does LCD stand for? And what does it really mean?

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2.

When you clear denominators from an equation like $\frac{2}{3}x+5=\frac{2}{7}\text{,}$ you will multiply by $21\text{.}$ At first, what are the two things that you multiply by $21\text{,}$ possibly requiring you to use some parentheses?

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3.

What is a proportional equation?

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Exercises 2.3.6 Exercises

Review and Warmup

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1.

Multiply: $\displaystyle{4\cdot \frac{2}{3} }$

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2.

Multiply: $\displaystyle{7\cdot \frac{3}{8} }$

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3.

Multiply: $\displaystyle{25\cdot\left( -{\frac{6}{5}} \right)}$

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4.

Multiply: $\displaystyle{30\cdot\left( -{\frac{7}{10}} \right)}$

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5.

Do the following multiplications.

$32 \cdot \frac{5}{8}$
$40 \cdot \frac{5}{8}$
$48 \cdot \frac{5}{8}$

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6.

Do the following multiplications.

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

Solving Linear Equations with Fractions

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7.

Solve the equation.

$\displaystyle{ {\frac{x}{10}+95}={2x} }$

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8.

Solve the equation.

$\displaystyle{ {\frac{r}{7}+68}={5r} }$

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9.

Solve the equation.

$\displaystyle{ {\frac{t}{4}+5}={9} }$

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10.

Solve the equation.

$\displaystyle{ {\frac{b}{9}+3}={9} }$

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11.

Solve the equation.

$\displaystyle{ {6-\frac{c}{5}} = {0} }$

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12.

Solve the equation.

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

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13.

Solve the equation.

$\displaystyle{ {-3} = {9-\frac{4C}{3}} }$

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14.

Solve the equation.

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

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15.

Solve the equation.

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

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16.

Solve the equation.

$\displaystyle{ {2x} = {\frac{5x}{8}+44} }$

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17.

Solve the equation.

$\displaystyle{ {80} = {{\frac{4}{9}}y+4y} }$

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18.

Solve the equation.

$\displaystyle{ {36} = {{\frac{2}{5}}t+2t} }$

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19.

Solve the equation.

$\displaystyle{ {78-{\frac{7}{8}}b} = {4b} }$

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20.

Solve the equation.

$\displaystyle{ {51-{\frac{5}{6}}c} = {2c} }$

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21.

Solve the equation.

$\displaystyle{ {3B} = {{\frac{2}{7}}B+1} }$

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22.

Solve the equation.

$\displaystyle{ {3C} = {{\frac{10}{9}}C+8} }$

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23.

Solve the equation.

$\displaystyle{ {{\frac{5}{4}}-7n}={2} }$

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24.

Solve the equation.

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

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25.

Solve the equation.

$\displaystyle{ {{\frac{3}{8}}-{\frac{1}{8}}x}={1} }$

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26.

Solve the equation.

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

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27.

Solve the equation.

$\displaystyle{ {\frac{2t}{9}-6}={-{\frac{74}{9}}} }$

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28.

Solve the equation.

$\displaystyle{ {\frac{4b}{7}-4}={-{\frac{68}{7}}} }$

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29.

Solve the equation.

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

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30.

Solve the equation.

$\displaystyle{ {{\frac{4}{9}}+{\frac{8}{9}}B}={3B} }$

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31.

Solve the equation.

$\displaystyle{ {\frac{4C}{7}-{\frac{30}{7}}}={-{\frac{2}{7}}C} }$

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32.

Solve the equation.

$\displaystyle{ {\frac{2n}{5}-{\frac{6}{5}}}={-{\frac{1}{5}}n} }$

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33.

Solve the equation.

$\displaystyle{ {\frac{10p}{9}+{\frac{7}{10}}}={p} }$

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34.

Solve the equation.

$\displaystyle{ {\frac{6x}{7}+{\frac{5}{6}}}={x} }$

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35.

Solve the equation.

$\displaystyle{ {\frac{4y}{3}-69}={-{\frac{5}{2}}y} }$

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36.

Solve the equation.

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

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37.

Solve the equation.

$\displaystyle{ {-{\frac{9}{10}}a+42}={\frac{3a}{20}} }$

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38.

Solve the equation.

$\displaystyle{ {-{\frac{5}{2}}c+19}={\frac{9c}{4}} }$

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39.

Solve the equation.

$\displaystyle{ {\frac{3B}{10}-10B}={{\frac{3}{20}}} }$

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40.

Solve the equation.

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

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41.

Solve the equation.

$\displaystyle{ {\frac{9n}{2}+{\frac{8}{7}}}={{\frac{1}{8}}n} }$

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42.

Solve the equation.

$\displaystyle{ {\frac{7p}{6}+{\frac{8}{5}}}={{\frac{1}{4}}p} }$

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43.

Solve the equation.

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

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44.

Solve the equation.

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

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45.

Solve the equation.

$\displaystyle{ {{\frac{7}{8}}}={\frac{t}{32}} }$

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46.

Solve the equation.

$\displaystyle{ {{\frac{9}{4}}}={\frac{a}{12}} }$

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47.

Solve the equation.

$\displaystyle{ {-\frac{c}{15}}={{\frac{10}{3}}} }$

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48.

Solve the equation.

$\displaystyle{ {-\frac{B}{35}}={{\frac{10}{7}}} }$

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49.

Solve the equation.

$\displaystyle{ {-\frac{C}{12}}={-{\frac{3}{4}}} }$

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50.

Solve the equation.

$\displaystyle{ {-\frac{n}{20}}={-{\frac{7}{10}}} }$

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51.

Solve the equation.

$\displaystyle{ {-{\frac{3}{7}}}={\frac{8p}{9}} }$

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52.

Solve the equation.

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

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53.

Solve the equation.

$\displaystyle{ {{\frac{9}{10}}}={\frac{y+3}{50}} }$

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54.

Solve the equation.

$\displaystyle{ {{\frac{5}{6}}}={\frac{t+1}{30}} }$

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55.

Solve the equation.

$\displaystyle{ \frac{7}{4} = \frac{a-7}{7} }$

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56.

Solve the equation.

$\displaystyle{ \frac{3}{10} = \frac{c-7}{5} }$

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57.

Solve the equation.

$\displaystyle{ {\frac{A-6}{4}}={\frac{A+2}{6}} }$

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58.

Solve the equation.

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

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59.

Solve the equation.

$\displaystyle{ {\frac{n+5}{6}-\frac{n-7}{12}}={{\frac{5}{3}}} }$

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60.

Solve the equation.

$\displaystyle{ {\frac{p+9}{4}-\frac{p-4}{8}}={{\frac{15}{8}}} }$

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61.

Solve the equation.

$\displaystyle{ {\frac{x}{3}-1}={\frac{x}{6}} }$

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62.

Solve the equation.

$\displaystyle{ {\frac{y}{7}-9}={\frac{y}{10}} }$

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63.

Solve the equation.

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

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64.

Solve the equation.

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

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65.

Solve the equation.

$\displaystyle{ {{\frac{4}{9}}c+{\frac{2}{9}}}={{\frac{7}{9}}c+{\frac{1}{3}}} }$

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66.

Solve the equation.

$\displaystyle{ {{\frac{5}{3}}A+{\frac{4}{3}}}={{\frac{8}{3}}A+2} }$

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67.

Solve the equation.

$\displaystyle{ {\frac{5C+8}{2}-\frac{4-C}{4}}={{\frac{3}{7}}} }$

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68.

Solve the equation.

$\displaystyle{ {\frac{3n+8}{4}-\frac{1-n}{8}}={{\frac{2}{3}}} }$

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69.

Solve the equation.

$\displaystyle{ {33}={\frac{p}{5}+\frac{p}{6}} }$

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70.

Solve the equation.

$\displaystyle{ {26}={\frac{x}{3}+\frac{x}{10}} }$

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71.

Solve the equation.

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

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72.

Solve the equation.

$\displaystyle{ {{\frac{4}{9}}t-1}={{\frac{3}{4}}t-4} }$

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73.

Solve the equation.

$\displaystyle{ {-{\frac{5}{2}}a-1}={{\frac{7}{4}}a-{\frac{3}{10}}} }$

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74.

Solve the equation.

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

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75.

Solve the equation.

$\displaystyle{ {-\frac{A}{5}+2}={-3} }$
$\displaystyle{ {\frac{-t}{5}+2}={-3} }$
$\displaystyle{ {\frac{y}{-5}+2}={-3} }$
$\displaystyle{ {\frac{-r}{-5}+2}={-3} }$

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76.

Solve the equation.

$\displaystyle{ {-\frac{C}{5}+3}={1} }$
$\displaystyle{ {\frac{-q}{5}+3}={1} }$
$\displaystyle{ {\frac{c}{-5}+3}={1} }$
$\displaystyle{ {\frac{-p}{-5}+3}={1} }$

Applications

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77.

Kimball is jogging in a straight line. He got a head start of $10$ meters from the starting line, and he ran $5$ meters every $8$ seconds. After how many seconds will Kimball be $30$ meters away from the starting line?

Kimball will be $30$ meters away from the starting line seconds since he started running.

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78.

Brent is jogging in a straight line. He started at a place $36$ meters from the starting line, and ran toward the starting line at the speed of $5$ meters every $7$ seconds. After how many seconds will Brent be $21$ meters away from the starting line?

Brent will be $21$ meters away from the starting line seconds since he started running.

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79.

Joseph had only ${\$7.00}$ in his piggy bank, and he decided to start saving more. He saves ${\$4.00}$ every $5$ days. After how many days will he have ${\$23.00}$ in the piggy bank?

Joseph will save ${\$23.00}$ in his piggy bank after days.

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80.

Brandon has saved ${\$41.00}$ in his piggy bank, and he decided to start spending them. He spends ${\$2.00}$ every $7$ days. After how many days will he have ${\$31.00}$ left in the piggy bank?

Brandon will have ${\$31.00}$ left in his piggy bank after days.

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81.

According to a salad recipe, each serving requires $2$ teaspoons of vegetable oil and $8$ teaspoons of vinegar. If $17$ teaspoons of vegetable oil were used, how many teaspoons of vinegar should be used?

If $17$ teaspoons of vegetable oil were used, teaspoons of vinegar should be used.

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82.

According to a salad recipe, each serving requires $5$ teaspoons of vegetable oil and $20$ teaspoons of vinegar. If $84$ teaspoons of vinegar were used, how many teaspoons of vegetable oil should be used?

If $84$ teaspoons of vinegar were used, teaspoons of vegetable oil should be used.

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83.

Jay makes $\$108$ every eight hours he works. How much will he make if he works twenty-four hours this week?

If Jay works twenty-four hours this week, he will make .

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84.

Casandra makes $\$198$ every twelve hours she works. How much will she make if she works thirty hours this week?

If Casandra works thirty hours this week, she will make .

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85.

A mutual fund consists of $63$ % stock and $37$ % bond. In other words, for each $63$ dollars of stock, there are $37$ dollars of bond. For a mutual fund with ${\$2{,}810.00}$ of stock, how many dollars of bond are there?

For a mutual fund with ${\$2{,}810.00}$ of stock, there are approximately of bond.

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86.

A mutual fund consists of $72$ % stock and $28$ % bond. In other words, for each $72$ dollars of stock, there are $28$ dollars of bond. For a mutual fund with ${\$2{,}460.00}$ of bond, how many dollars of stock are there?

For a mutual fund with ${\$2{,}460.00}$ of bond, there are approximately of stock.

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87.

Eileen jogs every day. Last month, she jogged $6.5$ hours for a total of $37.05$ miles. At this speed, if Eileen runs $31.5$ hours, how far can she run?

At this speed, Eileen can run in $31.5$ hours.

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88.

Donna jogs every day. Last month, she jogged $16.5$ hours for a total of $57.75$ miles. At this speed, how long would it take Donna to run $140$ miles?

At this speed, Donna can run ${140\ {\rm mi}}$ in .

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89.

Kimball purchased $5.3$ pounds of apples at the total cost of ${\$7.42}\text{.}$ If he purchases $6.6$ pounds of apples at this store, how much would it cost?

It would cost to purchase $6.6$ pounds of apples.

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90.

Alejandro purchased $3.1$ pounds of apples at the total cost of ${\$6.20}\text{.}$ If the price doesn’t change, how many pounds of apples can Alejandro purchase with ${\$12.20}\text{?}$

With ${\$12.20}\text{,}$ Alejandro can purchase of apples.

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91.

Dawn collected a total of $1411$ stamps over the past $17$ years. At this rate, how many stamps would she collect in $27$ years?

At this rate, Dawn would collect stamps in $27$ years.

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92.

Corey collected a total of $1330$ stamps over the past $14$ years. At this rate, how many years would it take he to collect $2090$ stamps?

At this rate, Corey can collect $2090$ stamps in years.

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93.

In a city, the owner of a house valued at $260$ thousand dollars needs to pay ${\$821.60}$ in property tax. At this tax rate, how much property tax should the owner pay if a house is valued at $650$ thousand dollars?

The owner of a $650$ -thousand-dollar house should pay in property tax.

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94.

In a city, the owner of a house valued at $480$ thousand dollars needs to pay ${\$2{,}299.20}$ in property tax. At this tax rate, if the owner of a house paid ${\$3{,}113.50}$ of property tax, how much is the house worth?

If the owner of a house paid ${\$3{,}113.50}$ of property tax, the house is worth thousand dollars.

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95.

To try to determine the health of the Rocky Mountain elk population in the Wenaha Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released $39$ Rocky Mountain elk. A week later, they returned and observed $42$ Rocky Mountain elk, $9$ of which had tags. Approximately how many Rocky Mountain elk are in the Wenaha Wildlife Area?

There are approximately elk in the wildlife area.

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96.

To try to determine the health of the black-tailed deer population in the Jewell Meadow Wildlife Area, the Oregon Department of Fish and Wildlife caught, tagged, and released $28$ black-tailed deer. A week later, they returned and observed $63$ black-tailed deer, $18$ of which had tags. Approximately how many black-tailed deer are in the Jewell Meadow Wildlife Area?

There are approximately deer in the wildlife area.

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97.

A restaurant used ${1015\ {\rm lb}}$ of vegetable oil in $25$ days. At this rate, how many pounds of vegetable oil will be used in $42$ days?

The restaurant will use of vegetable oil in $42$ days.

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98.

A restaurant used ${735.9\ {\rm lb}}$ of vegetable oil in $33$ days. At this rate, ${1271.1\ {\rm lb}}$ of oil will last how many days?

The restaurant will use ${1271.1\ {\rm lb}}$ of vegetable oil in days.

Solving Inequalities with Fractions

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