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Solution - Absolute value equations

Exact form: x=0,6
x=0 , 6

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

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

2. Solve the two equations for x

8 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3x+3=3

Subtract from both sides:

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

Simplify the arithmetic:

3x=33

Simplify the arithmetic:

3x=0

Divide both sides by the coefficient:

x=0

11 additional steps

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

Expand the parentheses:

(x+3)=2x-3

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+3=3

Subtract from both sides:

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

Simplify the arithmetic:

x=33

Simplify the arithmetic:

x=6

Multiply both sides by :

-x·-1=-6·-1

Remove the one(s):

x=-6·-1

Simplify the arithmetic:

x=6

3. List the solutions

x=0,6
(2 solution(s))

4. Graph

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