Linear Equations with Fractions – Examples and Practice Problems

One of the most important topics in algebra is the process of solving linear equations. In this article, we will look at an introduction to linear equations in general and then focus on linear equations with fractions.

We will look at several equations with answers to facilitate learning the processes to solve these types of equations.

ALGEBRA
linear equations with fractions

Relevant for…

Learning to solve linear equations with fractions.

See process

ALGEBRA
First degree equations with fractions exercises

Relevant for…

Learning to solve linear equations with fractions.

See process

What are linear equations?

Linear equations have the characteristic that they can be written in the form ax+b=c, where x is a variable or unknown and a, ~b and c are known numbers.

For an equation to be linear, it must contain only variables raised to the first power after being reduced to its simplest form. For example, the equations 2x+6=-9 and 3x-5=x-7 are linear equations.

The equation 2 {{x}^2}+6x-2-2{{x}^2}=5 does not appear to be a linear equation. However, we can see that when we write it in its simplest form, that is, when we combine like terms, the terms that contain {{x}^2} disappear and we have the equation 6x=7. Therefore, this equation is a linear equation.

All linear equations with one variable and fractions have exactly one solution.

EXAMPLES

The following are linear equations:

2x+3x+3=15

\frac{1}{2}x+3=\frac{3}{2}+2x

5{{x}^{2}}+2x-4=5{{x}^{2}}+10

4x+{{x}^{3}}-{{x}^{3}}=10


Rule of equations

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.

For example, we have the equation 4x+4=2x-10. If we want to, we can add 5 to both sides of the equation: 4x+4+5=2x-10+5.

Note that with this equation, it was not very helpful to have added 5 to both sides of the equation. However, we can use the same idea to apply operations in a way that simplifies the equation and we can find its solution.

Now we will look at a method that we can use to solve linear equations with fractions.


Method for solving linear equations with fractions

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.

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.

3. Apply different operations to completely isolate the x.

4. Verify the answer.


Linear equations with fractions – Examples with answers

EXAMPLE 1

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

Solution: 

1. Simplify: 

We remove the parentheses: \frac{1}{2}x+4x+8=14-2x+20.

We simplify the fractions by multiplying by 2: x+8x+16=28-4x+40.

Combine like terms: 9x+16=68-4x.

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

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

9x=52-4x

9x+4x=52-4x+4x

13x=52

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

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

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

2+16+8=14-8+20

26=26

This is true

Answer: x=4

EXAMPLE 2

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

Solution: 

1. Simplify: 

There are no parentheses.

Simplify the fractions: multiply by 6: 2x+24+3x=54.

Combine like terms: 5x+24=54.

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

5x+24-24=54-24

5x=30

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

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

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

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

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

2+4+3=9

9=9

This is true

Answer: x=6

EXAMPLE 3

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

Solution: 

1. Simplify: 

Remove the parentheses: \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: 25x+20=12x+150.

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

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

25x=12x+130

25x-12x=12x-12x+130

13x=130

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

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

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

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

27=27

This is true

Answer: x=10

EXAMPLE 4

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

Solution: 

1. Simplify: 

Remove the parentheses: \frac{1}{3}x-4x-8=10-2x+17.

Remove the fractions: x-12x-24=30-6x+51.

Combine like terms: -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

-11x=-6x+105

-11x+6x=-6x+105+6x

-5x=105

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

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

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

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

-7+76=10+42+17

69=69

This is true

Answer: x=-21


Linear equations with fractions – Practice problems

Solve the equation \frac{1}{3}x+2=2x-3.

Choose an answer






Solve the equation \frac{1}{2}x+2=\frac{1}{3}x+3.

Choose an answer






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

Choose an answer






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

Choose an answer







See also

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