# How to Solve Linear Equations with Fractions – Examples

To solve linear equations with fractions, we have to start by simplifying the equation to the simplest version. This means that we have to multiply by the least common denominator to simplify all fractions. Afterward, we can solve the simplified equation easily.

Here, we will learn how to solve linear equations with fractions. We will look at a step-by-step process and use it to solve some practice examples.

##### ALGEBRA

Relevant for…

Learning to solve linear equations with fractions.

See method

##### ALGEBRA

Relevant for…

Learning to solve linear equations with fractions.

See method

## Method for solving linear equations with fractions

Recall that linear equations have the form $latex ax+b=c$ when simplified, where $latex x$ is a variable or unknown and $latex a, b$ and $latex c$ are known numbers.

In addition, we know that it is possible to perform any operation (addition, subtraction, multiplication, division) on one side of the equation, as long as we perform the same operation on the other side of the equation.

Therefore, we can solve linear equations by applying operations to both sides of the equation in order to completely isolate the variable.

When we have fractions in the equation, we can multiply by the least common denominator to eliminate the fractions. Thus, we can follow the following method to solve quadratic equations with fractions.

Step 1. Simplify both sides of the equation:

• a. Remove the parentheses (using the distributive property) if there are any.
• b. Remove the fractions (multiply both sides by the least common multiple).
• c. Combine like terms.

Step 2. Isolate the variable on one side of the equation:

We apply addition and subtraction so that the variable is on only one side of the equation and the constants are on the other side.

Step 3. Apply different operations to completely isolate the x.

Step 4. Verify the answer.

## Linear equations with fractions – Examples with answers

### EXAMPLE 1

Find the value of $latex x$ in the equation $latex \frac{1}{2}x+4(x+2)=14-2x+20$.

1. Simplify:

• We remove the parentheses: $latex \frac{1}{2}x+4x+8=14-2x+20$.
• We simplify the fractions by multiplying by 2: $latex x+8x+16=28-4x+40$.
• Combine like terms: $latex 9x+16=68-4x$.

2. Isolate the variable: move the 16 to the right and the -4x to the left:

$latex 9x+16-16=68-4x-16$

$latex 9x=52-4x$

$latex 9x+4x=52-4x+4x$

$latex 13x=52$

3. Apply different operations to isolate the $latex x$: divide both sides by 13:

$latex \frac{13}{13}x=\frac{{52}}{13}$

$latex x=\frac{{52}}{13}=4$

4. Verify your solution: substitute the value in the original equation:

$$\frac{1}{2}(4)+4(4+2)=14-2(4)+20$$

$latex 2+16+8=14-8+20$

$latex 26=26$

This is true

Answer: $latex x=4$

### EXAMPLE 2

Find the value of $latex x$ in the equation $latex \frac{1}{3}x+4+\frac{1}{2}x=9$.

1. Simplify:

• There are no parentheses.
• Simplify the fractions: multiply by 6: $latex 2x+24+3x=54$.
• Combine like terms: $latex 5x+24=54$.

2. Isolate the variable: move the 24 to the right:

$latex 5x+24-24=54-24$

$latex 5x=30$

3. Apply different operations to isolate the $latex x$: divide both sides by 5:

$latex \frac{5}{5}x=\frac{{30}}{5}$

$latex x=\frac{{30}}{5}=6$

4. Verify your solution: substitute the value in the original equation:

$latex \frac{1}{3}(6)+4+\frac{1}{2}(6)=9$

$latex 2+4+3=9$

$latex 9=9$

This is true

Answer: $latex x=6$.

### EXAMPLE 3

Find the value of $latex x$ in the equation $$\frac{1}{2}x+2(x+1)=\frac{1}{5}x+4+x+11$$

1. Simplify:

• Remove the parentheses: $latex \frac{1}{2}x+2x+2=\frac{1}{5}x+4+x+11$.
• Remove the fractions: multiply by 10: $$5x+20x+20=2x+40+10x+110$$
• Combine like terms: $latex 25x+20=12x+150$.

2. Isolate the variable: move the 20 to the right and the 12x to the left:

$latex 25x+20-20=12x+150-20$

$latex 25x=12x+130$

$latex 25x-12x=12x-12x+130$

$latex 13x=130$

3. Apply operations to isolate the $latex x$: divide both sides by 13:

$latex \frac{13}{13}x=\frac{{130}}{13}$

$latex x=\frac{{130}}{13}=10$

4. Verify your solution: substitute the value in the original equation:

$$\frac{1}{2}(10)+2(10+1)=\frac{1}{5}(10)+4+10+11$$

$latex 5+2(11)=2+4+10+11$

$latex 27=27$

This is true

Answer: $latex x=10$.

### EXAMPLE 4

Find the value of $latex x$ in the equation $latex \frac{1}{3}x-4(x+2)=10-2x+17$.

1. Simplify:

• Remove the parentheses: $latex \frac{1}{3}x-4x-8=10-2x+17$.
• Remove the fractions: $latex x-12x-24=30-6x+51$.
• Combine like terms: $latex -11x-24=-6x+81$.

2. Isolate the variable: move the -24 to the right and the -6x to the left:

$$-11x-24+24=-6x+81+24$$

$latex -11x=-6x+105$

$latex -11x+6x=-6x+105+6x$

$latex -5x=105$

3. Apply operations to isolate the $latex x$: divide both sides by -5:

$latex \frac{-5}{-5}x=\frac{{105}}{-5}$

$latex x=\frac{{105}}{-5}=-21$

4. Verify your solution: substitute the value in the original equation:

$$\frac{1}{3}(-21)-4(-21+2)=10-2(-21)+17$$

$latex -7-4(-19)=10+42+17$

$latex -7+76=10+42+17$

$latex 69=69$

This is true

Answer: $latex x=-21$.

## Linear equations with fractions – Practice problems

#### Find the value of x in the equation $latex \frac{2}{5}x+5(x-4)=x+2$.  