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Solution - Absolute value equations

Exact form: x=-5,15
x=-5 , \frac{1}{5}
Decimal form: x=5,0.2
x=-5 , 0.2

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
|3x+2|=|2x3|
without the absolute value bars:

|x|=|y||3x+2|=|2x3|
x=+y(3x+2)=(2x3)
x=y(3x+2)=(2x3)
+x=y(3x+2)=(2x3)
x=y(3x+2)=(2x3)

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||3x+2|=|2x3|
x=+y , +x=y(3x+2)=(2x3)
x=y , x=y(3x+2)=(2x3)

2. Solve the two equations for x

7 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+2=3

Subtract from both sides:

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

Simplify the arithmetic:

x=32

Simplify the arithmetic:

x=5

10 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

5x+2=(-2x+3)+2x

Group like terms:

5x+2=(-2x+2x)+3

Simplify the arithmetic:

5x+2=3

Subtract from both sides:

(5x+2)-2=3-2

Simplify the arithmetic:

5x=32

Simplify the arithmetic:

5x=1

Divide both sides by :

(5x)5=15

Simplify the fraction:

x=15

3. List the solutions

x=-5,15
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x+2|
y=|2x3|
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.