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Solution - Absolute value equations

Exact form: x=3,3
x=-3 , 3

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|x+3|+|x+3|=0

Add |x+3| to both sides of the equation:

|x+3|+|x+3||x+3|=|x+3|

Simplify the arithmetic

|x+3|=|x+3|

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

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

3. Solve the two equations for x

12 additional steps

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

Expand the parentheses:

(x+3)=-x-3

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x+3=3

Subtract from both sides:

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

Simplify the arithmetic:

2x=33

Simplify the arithmetic:

2x=6

Divide both sides by :

(2x)2=-62

Simplify the fraction:

x=-62

Find the greatest common factor of the numerator and denominator:

x=(-3·2)(1·2)

Factor out and cancel the greatest common factor:

x=3

5 additional steps

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

NT_MSLUS_MAINSTEP_RESOLVE_DOUBLE_MINUS:

(x+3)=x+3

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

3=(x+3)-x

Group like terms:

3=(x-x)+3

Simplify the arithmetic:

3=3

4. List the solutions

x=3,3
(2 solution(s))

5. Graph

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