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Solution - Absolute value equations

Exact form: =5,2
=5 , 2

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
|+3|=|2x7|
without the absolute value bars:

|x|=|y||+3|=|2x7|
x=+y(+3)=(2x7)
x=y(+3)=(2x7)
+x=y(+3)=(2x7)
x=y(+3)=(2x7)

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||+3|=|2x7|
x=+y , +x=y(+3)=(2x7)
x=y , x=y(+3)=(2x7)

2. Solve the two equations for

7 additional steps

(3)=(2x-7)

Swap sides:

(2x-7)=(3)

Add to both sides:

(2x-7)+7=(3)+7

Simplify the arithmetic:

2x=(3)+7

Simplify the arithmetic:

2x=10

Divide both sides by :

(2x)2=102

Simplify the fraction:

x=102

Find the greatest common factor of the numerator and denominator:

x=(5·2)(1·2)

Factor out and cancel the greatest common factor:

x=5

10 additional steps

(3)=-(2x-7)

Expand the parentheses:

(3)=-2x+7

Swap sides:

-2x+7=(3)

Subtract from both sides:

(-2x+7)-7=(3)-7

Simplify the arithmetic:

-2x=(3)-7

Simplify the arithmetic:

2x=4

Divide both sides by :

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

Cancel out the negatives:

2x2=-4-2

Simplify the fraction:

x=-4-2

Cancel out the negatives:

x=42

Find the greatest common factor of the numerator and denominator:

x=(2·2)(1·2)

Factor out and cancel the greatest common factor:

x=2

3. List the solutions

=5,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|+3|
y=|2x7|
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.