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Solution - Absolute value equations

Exact form: x=12,14
x=\frac{1}{2} , \frac{1}{4}
Decimal form: x=0.5,0.25
x=0.5 , 0.25

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation without absolute value bars

Use the rules:
|x|=|y|x=±y and |x|=|y|±x=y
to write all four options of the equation
|5x2|=|x+1|
without the absolute value bars:

|x|=|y||5x2|=|x+1|
x=+y(5x2)=(x+1)
x=y(5x2)=(x+1)
+x=y(5x2)=(x+1)
x=y(5x2)=(x+1)

When simplified, equations x=+y and +x=y are the same and equations x=y and x=y are the same, so we end up with only 2 equations:

|x|=|y||5x2|=|x+1|
x=+y , +x=y(5x2)=(x+1)
x=y , x=y(5x2)=(x+1)

2. Solve the two equations for x

11 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

6x-2=(-x+1)+x

Group like terms:

6x-2=(-x+x)+1

Simplify the arithmetic:

6x2=1

Add to both sides:

(6x-2)+2=1+2

Simplify the arithmetic:

6x=1+2

Simplify the arithmetic:

6x=3

Divide both sides by :

(6x)6=36

Simplify the fraction:

x=36

Find the greatest common factor of the numerator and denominator:

x=(1·3)(2·3)

Factor out and cancel the greatest common factor:

x=12

10 additional steps

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

Expand the parentheses:

(5x-2)=x-1

Subtract from both sides:

(5x-2)-x=(x-1)-x

Group like terms:

(5x-x)-2=(x-1)-x

Simplify the arithmetic:

4x-2=(x-1)-x

Group like terms:

4x-2=(x-x)-1

Simplify the arithmetic:

4x2=1

Add to both sides:

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

Simplify the arithmetic:

4x=1+2

Simplify the arithmetic:

4x=1

Divide both sides by :

(4x)4=14

Simplify the fraction:

x=14

3. List the solutions

x=12,14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x2|
y=|x+1|
The equation is true where the two lines cross.

Why learn this

We encounter absolute values almost daily. For example: If you walk 3 miles to school, do you also walk minus 3 miles when you go back home? The answer is no because distances use absolute value. The absolute value of the distance between home and school is 3 miles, there or back.
In short, absolute values help us deal with concepts like distance, ranges of possible values, and deviation from a set value.