Linear Equations with Fractions - Examples and Practice Problems

Linear equations are equations that only have variables with a power of 1. To solve linear equation problems with fractions, we have to multiply the entire equation by the least common multiple to eliminate the fractions and then use the conventional method of solving first-degree equations.

In this article, we will look at several examples with answers to master this topic and we will also look at some practice problems and practice what we have learned.

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Exploring linear equations with fractions examples and practice problems.

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ALGEBRA

Relevant for…

Exploring linear equations with fractions examples and practice problems.

See examples

Summary of linear equations with fractions

Recall that, for an equation to be of the first degree, all the variables in the equation must have a maximum power of 1. For example, the equations $\frac{1}{3}x+2=6$ and $\frac{3}{2}x=\frac{1}{2}x+10$ are first degree equations with fractions. We can solve first degree equations with fractions with the following steps:

Step 1: Remove the fractions: We multiply the entire equation by the least common multiple to remove the fractions.

Step 2: Simplify: We remove the parentheses and other grouping signs and combine like terms.

Step 3: Solve for the variable: We use addition and subtraction to move the variable to only one side of the equation.

Step 4: Solve: We use multiplication and division to solve for the variable completely.

Examples with answers of linear equations with fractions

The following examples of linear equations with fractions have their respective solution. Therefore, you can follow the solving process and fully master solving these types of equations.

EXAMPLE 1

Solve the equation $\frac{x}{3}+4=6$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 3:

$\frac{x}{3}+4=6$

$x+4(3)=6(3)$

$x+12=18$

Step 2: Simplify: We have nothing to simplify.

Step 3: Solve for the variable: We subtract 12 from both sides of the equation:

$x+12-12=18-12$

$x=6$

Step 4: Solve: We have already got the answer:

$x=6$

EXAMPLE 2

Find the value of x in the equation $\frac{x+1}{2}+2=4$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 2:

$\frac{x+1}{2}+2=4$

$x+1+2(2)=4(2)$

$x+1+4=8$

Step 2: Simplify: We combine like terms:

$x+1+4=8$

$x+5=8$

Step 3: Solve for the variable: We subtract 5 from both sides:

$x+5-5=8-5$

$x=3$

Step 4: Solve: We have already got the answer:

$x=3$

EXAMPLE 3

Find the value of x in the equation $\frac{2x+5}{3}+2x=7$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 3:

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

$2x+5+2x(3)=7(3)$

$2x+5+6x=21$

Step 2: Simplify: We combine like terms:

$2x+5+6x=21$

$8x+5=21$

Step 3: Solve for the variable: We subtract 5 from both sides:

$8x+5-5=21-5$

$8x=16$

Step 4: Solve: We divide both sides by 8:

$\frac{8x}{8}=\frac{16}{8}$

$x=2$

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EXAMPLE 4

Find the value of x in the equation $\frac{3x-4}{4}+6=2x+10$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 4:

$\frac{3x-4}{4}+6=2x+10$

$3x-4+6(4)=2x(4)+10(4)$

$3x-4+24=8x+40$

Step 2: Simplify: We combine like terms:

$3x-4+24=8x+40$

$3x+20=8x+40$

Step 3: Solve for the variable: We subtract 20 and 8x from both sides of the equation:

$3x+20-20=8x+40-20$

$3x=8x+20$

$3x-8x=8x+20-8x$

$-5x=20$

Step 4: Solve: We divide both sides by -5:

$\frac{-5x}{-5}=\frac{20}{-5}$

$x=-4$

EXAMPLE 5

Solve the equation $\frac{t+5}{2}+5=\frac{t-6}{3}+10$ for t.

Solution

Step 1: Remove the fractions: We multiply the entire equation by 6:

$\frac{t+5}{2}+5=\frac{t-6}{3}+10$

$3(t+5)+5(6)=2(t-6)+10(6)$

$3(t+5)+30=2(t-6)+60$

Step 2: Simplify: We expand the parentheses and combine like terms:

$3(t+5)+30=2(t-6)+60$

$3t+15+30=2t-12+60$

$3t+45=2t+48$

Step 3: Solve for the variable: We subtract 45 and 2 t from both sides:

$3t+45-45=2t+48-45$

$3t=2t+3$

$3t-2t=2t+3-2t$

$t=3$

Step 4: Solve: We already got the answer:

$t=3$

EXAMPLE 6

Solve the equation $\frac{5x-10}{2}+5=2(2x-2)+1$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 2:

$\frac{5x-10}{2}+5=2(2x-2)+1$

$5x-10+5(2)=2(2)(2x-2)+1(2)$

$5x-10+10=4(2x-2)+2$

Step 2: Simplify: We remove the parentheses and combine like terms:

$5x-10+10=4(2x-2)+2$

$5x-10+10=8x-8+2$

$5x=8x-6$

Step 3: Solve for the variable: We subtract 8x from both sides of the equation:

$5x-8x=8x-6-8x$

$-3x=-6$

Step 4: Solve: We divide both sides by -3:

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

$x=2$

EXAMPLE 7

Find the value of z from the equation $\frac{2z+1}{3}+\frac{z-1}{2}=\frac{-3z-5}{2}-11$ .

Solution

Step 1: Remove the fractions: We multiply the entire equation by 6 to eliminate all fractions:

$\frac{2z+1}{3}+\frac{z-1}{2}=\frac{-3z-5}{2}-11$

$2(2z+1)+3(z-1)=3(-3z-5)-6(11)$

$2(2z+1)+3(z-1)=3(-3z-5)-66$

Step 2: Simplify: We remove the parentheses and combine like terms:

$2(2z+1)+3(z-1)=3(-3z-5)-66$

$4z+2+3z-3=-9z-15-66$

$7z-1=-9z-81$

Step 3: Solve for the variable: Add 1 and 9 z from both sides of the equation:

$7z-1+1=-9z-81+1$

$7z=-9z-80$

$7z+9z=-9z-80+9z$

$16z=-80$

Step 4: Solve: We divide both sides by 16:

$\frac{16z}{16}=\frac{-80}{16}$

$x=-5$

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Linear equations with fractions – Practice problems

The following practice problems can be solved to practice the process of solving linear equations with fractions. If you have trouble solving these problems, you can carefully study the solved examples shown above.

What is the value of x in the equation $\frac{2x}{5}+5=7$ .

Choose an answer

$x=2$

$x=3$

$x=4$

$x=5$

Solve the equation $\frac{3x-6}{6}-4=-3$ .

Choose an answer

$x=2$

$x=-2$

$x=4$

$x=-4$

Find the value of x in the equation $\frac{5x+2}{4}+\frac{3x}{2}=-5$ .

Choose an answer

$x=-3$

$x=-2$

$x=2$

$x=3$

Solve the equation $3x-\frac{2x-10}{4}=2(x-2)+9$ .

Choose an answer

$x=0$

$x=3$

$x=5$

$x=7$

Solve the equation $\frac{4x+1}{3}+4=\frac{x+3}{2}+2(x+2)$ .

Choose an answer

$x=-1$

$x=-2$

$x=1$

$x=2$

Linear Equations with Fractions – Examples and Practice Problems

ALGEBRA

ALGEBRA

Summary of linear equations with fractions

Examples with answers of linear equations with fractions

EXAMPLE 1

Solution

EXAMPLE 2

Solution

EXAMPLE 3

Solution

EXAMPLE 4

Solution

EXAMPLE 5

Solution

EXAMPLE 6

Solution

EXAMPLE 7

Solution

Linear equations with fractions – Practice problems

What is the value of x in the equation $\frac{2x}{5}+5=7$ .

Solve the equation $\frac{3x-6}{6}-4=-3$ .

Find the value of x in the equation $\frac{5x+2}{4}+\frac{3x}{2}=-5$ .

Solve the equation $3x-\frac{2x-10}{4}=2(x-2)+9$ .

Solve the equation $\frac{4x+1}{3}+4=\frac{x+3}{2}+2(x+2)$ .

See also

Learn mathematics with our additional resources in different topics

ALGEBRA

ALGEBRA

Summary of linear equations with fractions

Examples with answers of linear equations with fractions

EXAMPLE 1

Solution

EXAMPLE 2

Solution

EXAMPLE 3

Solution

EXAMPLE 4

Solution

EXAMPLE 5

Solution

EXAMPLE 6

Solution

EXAMPLE 7

Solution

Linear equations with fractions – Practice problems

What is the value of x in the equation .

Solve the equation .

Find the value of x in the equation .

Solve the equation .

Solve the equation .

See also

Learn mathematics with our additional resources in different topics

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What is the value of x in the equation $\frac{2x}{5}+5=7$ .

Solve the equation $\frac{3x-6}{6}-4=-3$ .

Find the value of x in the equation $\frac{5x+2}{4}+\frac{3x}{2}=-5$ .

Solve the equation $3x-\frac{2x-10}{4}=2(x-2)+9$ .

Solve the equation $\frac{4x+1}{3}+4=\frac{x+3}{2}+2(x+2)$ .