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Solution - Absolute value equations

Exact form: x=5,-32
x=5 , -\frac{3}{2}
Mixed number form: x=5,-112
x=5 , -1\frac{1}{2}
Decimal form: x=5,1.5
x=5 , -1.5

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
|3x2|=|x+8|
without the absolute value bars:

|x|=|y||3x2|=|x+8|
x=+y(3x2)=(x+8)
x=y(3x2)=(x+8)
+x=y(3x2)=(x+8)
x=y(3x2)=(x+8)

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||3x2|=|x+8|
x=+y , +x=y(3x2)=(x+8)
x=y , x=y(3x2)=(x+8)

2. Solve the two equations for x

11 additional steps

(3x-2)=(x+8)

Subtract from both sides:

(3x-2)-x=(x+8)-x

Group like terms:

(3x-x)-2=(x+8)-x

Simplify the arithmetic:

2x-2=(x+8)-x

Group like terms:

2x-2=(x-x)+8

Simplify the arithmetic:

2x2=8

Add to both sides:

(2x-2)+2=8+2

Simplify the arithmetic:

2x=8+2

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

12 additional steps

(3x-2)=-(x+8)

Expand the parentheses:

(3x-2)=-x-8

Add to both sides:

(3x-2)+x=(-x-8)+x

Group like terms:

(3x+x)-2=(-x-8)+x

Simplify the arithmetic:

4x-2=(-x-8)+x

Group like terms:

4x-2=(-x+x)-8

Simplify the arithmetic:

4x2=8

Add to both sides:

(4x-2)+2=-8+2

Simplify the arithmetic:

4x=8+2

Simplify the arithmetic:

4x=6

Divide both sides by :

(4x)4=-64

Simplify the fraction:

x=-64

Find the greatest common factor of the numerator and denominator:

x=(-3·2)(2·2)

Factor out and cancel the greatest common factor:

x=-32

3. List the solutions

x=5,-32
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x2|
y=|x+8|
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.