1.7 Equations with Fractions

To solve equations that contain fractions, we can use the same rules of algebra that we have been using in previous lessons.

Example 1

Solve: \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}=\frac{1}{4}

Table 1.7.1
Steps Solution
Subtract \frac{1}{2} from both sides, then simplify the left side. \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}=\frac{1}{4}

\frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}-\frac{1}{2}=\frac{1}{4}-\frac{1}{2}

\frac{1}{8}\phantom{\rule{0.1em}{0ex}}x=\frac{1}{4}-\frac{1}{2}
Find the common denominator for the constant terms, and rewrite fractions as necessary. \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x=\frac{1}{4}-\frac{2}{4}
Subtract. \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x=-\frac{1}{4}
Multiply both sides by the reciprocal of \frac{1}{8}. \frac{8}{1} \bullet \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x=\frac{8}{1} \bullet -\frac{1}{4}
Simpify. x = –2

This method works fine, but it can be complicated to work with so many fractions. We will now cover an alternate method that can be used to solve equations with fractions. This alternate method involves eliminating the fractions.

Table 1.7.2
To solve equations with fractional coefficients by clearing the fractions:
Step 1: Find the least common denominator (LCD) of all the fractions in the equation.

Step 2: Multiply both sides of the equation by the LCD.

Step 3: Simplify each side to clear the fractions.

Step 4: Solve the equation.

We will now solve the same question using the new method.

Example 2

Solve: \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}=\frac{1}{4}

Table 1.7.3
Steps Solution
Find the least common denominator of all the fractions in the equation. \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}=\frac{1}{4}

LCD = 8
Multiply both sides of the equation by the LCD. 8\left(\frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+\frac{1}{2}\right)=8\left(\frac{1}{4}\right)
Distribute the 8, and simplify each side to clear the fractions. 8 \bullet \frac{1}{8}\phantom{\rule{0.1em}{0ex}}x+8 \bullet \frac{1}{2}=8 \bullet \frac{1}{4}

x+4=2
Solve the equation. x+4-4=2-4

x = -2
Practice 1

Solve.

a) \frac{x}{4}+\frac{1}{2}=\frac{5}{8}

 

 

 

 

 b) \frac{1}{6}\phantom{\rule{0.1em}{0ex}}y-\frac{1}{3}=\frac{1}{6}

 

 

 

 
c) a+\frac{3}{4}=\frac{3}{8}\phantom{\rule{0.1em}{0ex}}a-\frac{1}{2}

 

 

 

 

 

 

 d) 7=\frac{x}{2}+\frac{3x}{4}-\frac{2x}{3}

 

 

 

 

 

 
e) 1=\frac{1}{2}\left(4x+2\right)

 

 

 

 

 

 f) \frac{1}{2}\left(y-5\right)=\frac{1}{4}\left(y-1\right)

 

 

 

 

 

Homework 

  1. Solve.

    a) x+\frac{1}{3}=\frac{x}{6}-\frac{1}{2}

     

     

     

     

     

     

    		 b) -1=\frac{1}{2}\phantom{\rule{0.1em}{0ex}}u+\frac{1}{4}\phantom{\rule{0.1em}{0ex}}u-\frac{2}{3}\phantom{\rule{0.1em}{0ex}}u

     

     

     

     

     
    c) 8=\frac{9q+6}{3}

     

     

     

     

     

     

    		 d) \frac{1}{2}\left(m-3\right)=\frac{1}{4}\left(m-7\right)

     

     

     

     

     

     
    e) \frac{x}{4}-\frac{1}{2}=-\frac{3}{4}

     

     

     

     

     

     

    		 f) \frac{5}{6}\phantom{\rule{0.1em}{0ex}}y-\frac{2}{3}=-\frac{3}{2}

     

     

     

     

     

     
    g) \frac{1}{2}\phantom{\rule{0.1em}{0ex}}a+\frac{3}{8}=\frac{3}{4}

     

     

     

     

     

     

    		 h) 2=\frac{1}{3}\phantom{\rule{0.1em}{0ex}}x-\frac{1}{2}\phantom{\rule{0.1em}{0ex}}x+\frac{2}{3}\phantom{\rule{0.1em}{0ex}}x

     

     

     

     

     

     
    i) \frac{m}{4}-\frac{4m}{5}+\frac{1m}{2}=-1

     

     

     

     

     

     

    		 j) x+\frac{1}{2}=\frac{2x}{3}-\frac{1}{2}

     

     

     

     

     

     
    k) \frac{1}{3}\phantom{\rule{0.1em}{0ex}}w+\frac{5}{4}=w-\frac{1}{4}

     

     

     

     

     

     

    		 l) \frac{x}{2}-\frac{1}{4}=\frac{1x}{12}+\frac{1}{6}

     

     

     

     

     

     
    m) \frac{1}{3}\phantom{\rule{0.1em}{0ex}}b+\frac{1}{5}=\frac{2}{5}\phantom{\rule{0.1em}{0ex}}b-\frac{3}{5}

     

     

     

     

     

     

    		 n) 1=\frac{1}{6}\left(12x-6\right)

     

     

     

     

     

     
    o) \frac{1}{4}\left(p-7\right)=\frac{1}{3}\left(p+5\right)

     

     

     

     

     

     

     

    		 p) \frac{1}{2}\left(x+4\right)=\frac{3}{4}

     

     

     

     

     

     

     

  2. Explain how to find the least common denominator of \frac{3}{8},\phantom{\rule{0.2em}{0ex}}\frac{1}{6},\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}\frac{2}{3}.

     

     

     

     

  3. If an equation has fractions only on one side, why do you have to multiply both sides of the equation by the LCD?

     

     

     

     

  4. Solve for x: \frac{3x}{4}-\frac{5}{2}=\frac{x}{2}+\frac{7}{4}

    a) x = –32
    b) x = 50
    c) x = 17
    d) x = –12

  5. Solve for x: \frac{3}{5}\phantom{\rule{0.1em}{0ex}}x-\frac{2}{3}=\frac{1}{2}\phantom{\rule{0.1em}{0ex}}x+\frac{4}{15}

    a) x = \frac{28}{3}

    b) x = \frac{1}{2}

    c) x = \frac{15}{9}

    d) x = \frac{5}{8}

Answers

1.

a) –1 b) –12 c) 2 d) –1 e) –1 f) –1 g) \frac{3}{4} h) 4
i) 20 j) –3 k) \frac{9}{4} l) 1 m) 12 n) 1 o) –41 p) -\frac{5}{2}
2. Write out the multiples of 8, 6, and 3. Then find the lowest multiple that is common among them. 3. This ensures that the equation stays equal, or balanced.
4. c 5. a

Attribution

Unless otherwise indicated, material on this page has been adapted from the following resource:

Mazur, I. (2021). Introductory algebra. BCcampus. https://collection.bccampus.ca/textbooks/intermediate-algebra-bccampus-412/, licensed under CC BY 4.0

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Math 10C Workbook Copyright © 2026 by Leanne Thompson is licensed under a Creative Commons Attribution-NonCommercial 4.0 International License, except where otherwise noted.

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