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Solution - Absolute value equations

Exact form: x=-12
x=-\frac{1}{2}
Decimal form: x=0.5
x=-0.5

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
|x+2|=|x+3|
without the absolute value bars:

|x|=|y||x+2|=|x+3|
x=+y(x+2)=(x+3)
x=y(x+2)=(x+3)
+x=y(x+2)=(x+3)
x=y(x+2)=(x+3)

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||x+2|=|x+3|
x=+y , +x=y(x+2)=(x+3)
x=y , x=y(x+2)=(x+3)

2. Solve the two equations for x

11 additional steps

(-x+2)=(x+3)

Subtract from both sides:

(-x+2)-x=(x+3)-x

Group like terms:

(-x-x)+2=(x+3)-x

Simplify the arithmetic:

-2x+2=(x+3)-x

Group like terms:

-2x+2=(x-x)+3

Simplify the arithmetic:

2x+2=3

Subtract from both sides:

(-2x+2)-2=3-2

Simplify the arithmetic:

2x=32

Simplify the arithmetic:

2x=1

Divide both sides by :

(-2x)-2=1-2

Cancel out the negatives:

2x2=1-2

Simplify the fraction:

x=1-2

Move the negative sign from the denominator to the numerator:

x=-12

6 additional steps

(-x+2)=-(x+3)

Expand the parentheses:

(-x+2)=-x-3

Add to both sides:

(-x+2)+x=(-x-3)+x

Group like terms:

(-x+x)+2=(-x-3)+x

Simplify the arithmetic:

2=(-x-3)+x

Group like terms:

2=(-x+x)-3

Simplify the arithmetic:

2=3

The statement is false:

2=3

The equation is false so it has no solution.

3. List the solutions

x=-12
(1 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x+2|
y=|x+3|
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.