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Solution - Absolute value equations

Exact form: x=-3,-53
x=-3 , -\frac{5}{3}
Mixed number form: x=-3,-123
x=-3 , -1\frac{2}{3}
Decimal form: x=3,1.667
x=-3 , -1.667

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
|4x+8|=|2x+2|
without the absolute value bars:

|x|=|y||4x+8|=|2x+2|
x=+y(4x+8)=(2x+2)
x=y(4x+8)=(2x+2)
+x=y(4x+8)=(2x+2)
x=y(4x+8)=(2x+2)

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||4x+8|=|2x+2|
x=+y , +x=y(4x+8)=(2x+2)
x=y , x=y(4x+8)=(2x+2)

2. Solve the two equations for x

11 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x+8=2

Subtract from both sides:

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

Simplify the arithmetic:

2x=28

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

12 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

6x+8=(-2x-2)+2x

Group like terms:

6x+8=(-2x+2x)-2

Simplify the arithmetic:

6x+8=2

Subtract from both sides:

(6x+8)-8=-2-8

Simplify the arithmetic:

6x=28

Simplify the arithmetic:

6x=10

Divide both sides by :

(6x)6=-106

Simplify the fraction:

x=-106

Find the greatest common factor of the numerator and denominator:

x=(-5·2)(3·2)

Factor out and cancel the greatest common factor:

x=-53

3. List the solutions

x=-3,-53
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x+8|
y=|2x+2|
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.