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# 2.5: Equations with Fractions

Difficulty Level: At Grade Created by: CK-12

In this year’s student election for president, there were two candidates. The winner received 13\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?

### Guidance

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 an added step of multiplying and dividing the equation by the numerator and denominator in order to solve for the variable. As well, if your fractions do not have the same denominator, you must first find the least common denominator (LCD) before combining like terms or combining constants.

#### Example A

Solve: 13t+5=1\begin{align*}\frac{1}{3}t+5=-1\end{align*}.

Solution:

13t+513t+5513t(3)13tt=1=15=6=6(3)=18(Subtract 5 from both sides to isolate the variable)(Simplify)(Multiply both sides by the denominator (3) in the fraction)(Simplify)\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*}

#### Example B

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

Solution:

\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*}y = 2\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*}

#### Example C

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

Solution:

\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*}

#### Concept Problem Revisited

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 need to get only one variable into 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.

### Vocabulary

Fraction
A fraction is a part of a whole consisting of a numerator divided by a denominator. For example, if a pizza is cut into eight slices and you ate 3 slices, you would have eaten \begin{align*}\frac{3}{8}\end{align*} of the pizza. \begin{align*}\frac{3}{8}\end{align*} is a fraction with 3 being the numerator and 8 being the denominator.
Least Common Denominator
The least common denominator or lowest common denominator is the smallest number that all of the denominators (or the bottom numbers) can be divided into evenly. For example with the fractions \begin{align*}\frac{1}{2}\end{align*} and \begin{align*}\frac{1}{3}\end{align*}, the smallest number that both 2 and 3 will divide into evenly is 6. Therefore the least common denominator is 6.

### Guided Practice

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

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

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

1.

\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*}

2.

\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 numberator } ({\color{red}2}) \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*}

3.

\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*}

In order to simplify further, you need to make the fractions all have the same denominator. In order to do this you need to find the least common denominator (LCD). If you look at the last equation \begin{align*}\left(\frac{1}{4}w-\frac{2}{3}w=3 \right)\end{align*} you will see that the denominators are 4, 3, and 1. The equation can really be seen as \begin{align*}\left(\frac{1}{{\color{blue}4}}w-\frac{2}{{\color{blue}3}}w=\frac{3}{{\color{blue}1}}\right)\end{align*}.

So what is the least common denominator for the numbers 1, 3, and 4? Well all three numbers will divide into 12 evenly \begin{align*}({\color{blue}4} \times 3 = 12, {\color{blue}3} \times 4 = 12\end{align*}, and \begin{align*}{\color{blue}1} \times 12 = 12)\end{align*}. Therefore you need to multiply \begin{align*}\frac{1}{{\color{blue}4}}w\end{align*} by \begin{align*}\frac{3}{3},-\frac{2}{{\color{blue}3}}w\end{align*}, by \begin{align*}\frac{4}{4}\end{align*}, and \begin{align*}\frac{3}{{\color{blue}1}}\end{align*} by \begin{align*}\frac{12}{12}\end{align*} in order to carry on with this problem and solve for the variable. Notice that by multiplying by \begin{align*}\frac{3}{3}\end{align*}, you are really just multiplying by 1!

\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*}

### Practice

Solve for the variable in each of the following problems.

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*}

Solve for the variable in each of the following problems.

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*}

Solve for the variable in each of the following problems.

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*}

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Dec 19, 2012
Apr 29, 2014

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