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# 2.5: Equations with Fractions

Difficulty Level: Advanced Created by: CK-12
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Practice Equations with Fractions
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In this year’s student election for president, there were two candidates. The winner received \begin{align*}\frac{1}{3}\end{align*} more votes than the loser. If there were 584 votes cast for president, how many votes did each of the two candidates receive?

### Equations with Fractions

When introducing fractions into an equation, the same rules for solving any equation apply. You need to keep the equations in balance by adding, subtracting, multiplying, or dividing on both sides of the equals sign in order to isolate the variable. The goal still remains to get your variable alone on one side of the equals sign with your constant terms on the other in order to solve for this variable.

With fractions, there is sometimes an added step of multiplying and dividing the equation by the numerator and denominator in order to solve for the variable. Or, if there are multiple fractions that do not have the same denominator, you must first find the least common denominator (LCD) before combining like terms.

#### Solve for t

Solve: \begin{align*}\frac{1}{3}t+5=-1\end{align*}.

\begin{align*}\frac{1}{3}t+5 &= -1\\ \frac{1}{3}t+5 {\color{red}-5} &= -1 {\color{red}-5} && ( \text{Subtract} \ 5 \ \text{from both sides to isolate the variable})\\ \frac{1}{3}t &= -6 && ( \text{Simplify})\\ ({\color{red}\cancel{3}}) \frac{1}{\cancel{3}}t &= -6 ({\color{red}3}) && ( \text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ t &= -18 && ( \text{Simplify})\end{align*}

Therefore \begin{align*}t = -18\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{1}{3}t+5 &= -1\\ \frac{1}{3} ({\color{red}-18})+5 &= -1\\ -6+5 &= -1\\ -1 &= -1 \ \ \end{align*}

#### Solve for x

Solve: \begin{align*}\frac{3}{4}x-3=2\end{align*}.

\begin{align*}\frac{3}{4}x-3 &= 2\\ \frac{3}{4}x-3 {\color{red}+3} &= 2 {\color{red}+3} && ( \text{Add} \ 3 \ \text{to both sides to isolate the variable})\\ \frac{3}{4}x &= 5 && ( \text{Simplify})\\ ({\color{red}\cancel{4}}) \frac{3}{4}x &= 5({\color{red}4})&& ( \text{Multiply both sides by the denominator} \ ({\color{red}4}) \ \text{in the fraction)}\\ 3x &= 20 && ( \text{Simplify})\\ \frac{\cancel{3} x}{{\color{red}\cancel{3}}} &= \frac{20}{{\color{red}3}}\\ x &= \frac{20}{3}\end{align*}

Therefore \begin{align*}x=\frac{20}{3}\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{3}{4} x-3 &= 2\\ \frac{3}{4} \left(\frac{20}{3}\right)-3 &= 2\\ \frac{20}{4}-3 &= 2\\ 5-3 &= 2\\ 2 &= 2 \ \ \end{align*}

#### Solve for x

Solve: \begin{align*}\frac{2}{5}x-4=-\frac{1}{5}x+8\end{align*}.

\begin{align*}\frac{2}{5} x-4 &= -\frac{1}{5}x+8\\ \frac{2}{5}x {\color{red}+\frac{1}{5}x}-4 &= -\frac{1}{5}x {\color{red}+\frac{1}{5}x}+8 && ( \text{Add} \ \frac{1}{5}x \ \text{to both sides of the equal sign to combine variables})\\ \frac{3}{5} x-4 &= 8 && ( \text{Simplify})\\ \frac{3}{5}x-4 {\color{red}+4} &= 8 {\color{red}+4} && ( \text{Add} \ 4 \ \text{to both sides of the equation to isolate the variable})\\ \frac{3}{5}x &= 12 && (\text{Simplify})\\ ({\color{red}\cancel{5}}) \frac{3}{\cancel{5}}x &= 12({\color{red}5}) && (\text{Multiply both sides by the denominator } ({\color{red}5}) \text{ in the fraction})\\ 3x &= 60 && (\text{Simplify})\\ \frac{\cancel{3}x}{\cancel{3}} &= \frac{60}{3} && ( \text{Divide both sides by the numerator } ({\color{red}3}) \text{ in the fraction})\\ x &= 20 && ( \text{Simplify})\end{align*}

Therefore \begin{align*}x = 20\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{2}{5}x-4 &= -\frac{1}{5}x+8\\ \frac{2}{5} ({\color{red}20})-4 &=- \frac{1}{5} ({\color{red}20})+8\\ \frac{40}{5}-4 &=-\frac{20}{5}+8\\ 8-4 &= -4+8\\ 4 &= 4 \ \ \end{align*}

### Examples

#### Example 1

Earlier, you were given a problem about a student election for president.

In this year’s student election for president, there were two candidates. The winner received \begin{align*}\frac{1}{3}\end{align*} more votes. If there were 584 votes cast for president, how many votes did each of the two candidates receive?

Let \begin{align*}x =\end{align*} votes for candidate 1 (the winner)

Let \begin{align*}y =\end{align*} votes for candidate 2

\begin{align*}x + y = 584\end{align*}

You must have only one variable in the equation in order to solve it. Let’s look at another relationship from the problem.

\begin{align*}x &= y + \frac{1}{3} y && (\text{Candidate} \ 1 \ \text{received} \ \frac{1}{3} \ \text{more votes than candidate} \ 2)\\ x &= \frac{3}{3} y + \frac{1}{3} y && ( \text{Make denominator common for both} \ y \ \text{variables})\\ x &= \frac{4}{3} y && ( \text{Simplify})\end{align*}

Now substitute into the original problem.

\begin{align*}\frac{4}{3} y + y &= 584 && (\text{Substitute for} \ x \ \text{into the equation})\\ \frac{4}{3} y + \frac{3}{3} y &= 584 && (\text{Make denominator common for both} \ y \ \text{variables})\\ \frac{7}{3} y &= 584 && (\text{Combine like terms})\\ {\color{red}(\cancel{3})} \frac{7}{\cancel{3}} y &= 584 {\color{red}(3)} && (\text{Multiply both sides by the denominator in the fraction})\\ 7y &= 1752 && (\text{Simplify})\\ \frac{\cancel{7} y}{{\color{red}\cancel{7}}} &= \frac{1752}{{\color{red}7}} && (\text{Divide both sides by the numerator in the fraction})\\ y &= 250.28 && (\text{Simplify})\end{align*}

So candidate 2 received 250 votes. Candidate 1 would then receive \begin{align*}584 - 250 = 334\end{align*} votes.

#### Example 2

Solve for x: \begin{align*}\frac{2}{3}x=12\end{align*}.

\begin{align*}\frac{2}{3}x &= 12\\ ({\color{red}\cancel{3}}) \frac{2}{3}x &= 12 ({\color{red}3}) && (\text{Multiply both sides by the denominator} \ ({\color{red}3}) \ \text{in the fraction})\\ 2x &= 36 && (\text{Simplify})\\ \frac{\cancel{2} x}{{\color{red}\cancel{2}}} &= \frac{36}{{\color{red}2}} && (\text{Divide both sides by the numerator} \ ({\color{red}2}) \ \text{in the fraction})\\ x &= 18 && (\text{Simplify})\end{align*}

Therefore \begin{align*}x = 18\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{2}{3}x &= 12\\ \frac{2}{3} ({\color{red}18}) &= 12\\ \frac{36}{3} &= 12\\ 12 &= 12 \ \ \end{align*}

#### Example 3

Solve for x: \begin{align*}\frac{3}{4}x-5=19\end{align*}.

\begin{align*}\frac{3}{4}x-5 &= 19\\ \frac{3}{4}x-5 {\color{red}+5} &= 19 {\color{red}+5} && (\text{Add} \ 5 \ \text{to both sides of the equal sign to isolate the variable})\\ \frac{3}{4}x &= 24 && (\text{Simplify})\\ ({\color{red}\cancel{4}}) \frac{3}{\cancel{4}}x &= 24({\color{red}4}) && (\text{Multiply both sides by the denominator} ({\color{red}4}) \ \text{in the fraction})\\ 3x &= 96 && (\text{Simplify})\\ \frac{\cancel{3} x}{{\color{red}\cancel{3}}}&=\frac{96}{{\color{red}3}} && (\text{Divide both sides by numerator } ({\color{red}3}) \text{ in the fraction})\\ x &= 32 && (\text{Simplify})\end{align*}

Therefore \begin{align*}x = 32\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{3}{4}x-5 &= 19\\ \frac{3}{4} ({\color{red}32})-5 &= 19\\ \frac{96}{4}-5 &= 19\\ 24-5 &= 19\\ 19 &= 19 \ \ \end{align*}

#### Example 4

Solve for x: \begin{align*}\frac{1}{4}w-3=\frac{2}{3}w\end{align*}.

\begin{align*}\frac{1}{4}w-3 &= \frac{2}{3}w\\ \frac{1}{4}w-3 {\color{red}+3} &= \frac{2}{3}w {\color{red}+3} && (\text{Add} \ 3 \ \text{to both sides of the equal sign to start})\\ \frac{1}{4}w &= \frac{2}{3}w+3 && (\text{Simplify})\\ \frac{1}{4}w {\color{red}-\frac{2}{3}w} &= \frac{2}{3}w {\color{red}-\frac{2}{3}w}+3 && (\text{Subtract} \ \frac{2}{3}w \ \text{from both sides of the equal sign to get variables on same side})\\ \frac{1}{4}w-\frac{2}{3}w &= 3 && (\text{Simplify})\end{align*}

\begin{align*}{\color{magenta}\left(\frac{3}{3}\right)} \frac{1}{4}w- {\color{magenta}\left(\frac{4}{4}\right)} \frac{2}{3}w &= {\color{magenta} \left(\frac{12}{12}\right)} 3 && (\text{Multiply by the LCD})\\ \frac{3}{12}w-\frac{8}{12}w &= \frac{36}{12} && (\text{Simplify})\end{align*}

Since all the denominators are the same (12), we can simplify further:

\begin{align*}3w-8w &= 36 && (\text{Combine like terms})\\ -5w &= 36 && (\text{Simplify})\\ \frac{-5w}{{\color{red}-5}} &= \frac{36}{{\color{red}-5}} && (\text{Divide by} \ -5 \ \text{to solve for the variable})\\ w &= -\frac{36}{5} && (\text{Simplify})\end{align*}

Therefore \begin{align*}w = -\frac{36}{5}\end{align*}.

\begin{align*}\text{Check:} &\\ \frac{1}{4}w-3 &= \frac{2}{3}w\\ \frac{1}{4} \left({\color{red}\frac{-36}{5}} \right)-3 &= \frac{2}{3} \left({\color{red}\frac{-36}{5}}\right)\\ \frac{-36}{20} -3 &= \frac{-72}{15}\\ \frac{-108}{60}-\frac{180}{60} &= \frac{-288}{60}\\ \frac{-288}{60} &= \frac{-288}{60} \ \ \end{align*}

### Review

Solve for the variable in each of the following equations.

1. \begin{align*}\frac{1}{3}p=5\end{align*}
2. \begin{align*}\frac{3}{7}j=8\end{align*}
3. \begin{align*}\frac{2}{5}b+4=6\end{align*}
4. \begin{align*}\frac{2}{3}x-2=1\end{align*}
5. \begin{align*}\frac{1}{3}x+3=-3\end{align*}
1. \begin{align*}\frac{1}{8}k+\frac{2}{3}=5\end{align*}
2. \begin{align*}\frac{1}{6}c+\frac{1}{3}=-2\end{align*}
3. \begin{align*}\frac{4}{5}x+3=\frac{2}{3}\end{align*}
4. \begin{align*}\frac{3}{4}x-\frac{2}{5}=\frac{1}{2}\end{align*}
5. \begin{align*}\frac{1}{4}t+\frac{2}{3}=\frac{1}{2}\end{align*}
1. \begin{align*}\frac{1}{3}x+\frac{1}{4}x=1\end{align*}
2. \begin{align*}\frac{1}{5}d+\frac{2}{3}d=\frac{5}{3}\end{align*}
3. \begin{align*}\frac{1}{2}x-1=\frac{1}{3}x\end{align*}
4. \begin{align*}\frac{1}{3}x-\frac{1}{2}=\frac{3}{4}x\end{align*}
5. \begin{align*}\frac{2}{3}j-\frac{1}{2}=\frac{3}{4}j+\frac{1}{3}\end{align*}

To see the Review answers, open this PDF file and look for section 2.5.

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Color Highlighted Text Notes

### Vocabulary Language: English

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.

fraction

A fraction is a part of a whole. A fraction is written mathematically as one value on top of another, separated by a fraction bar. It is also called a rational number.

Least Common Denominator

The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.

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