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Solution - Absolute value equations

Exact form: x=32,13
x=\frac{3}{2} , \frac{1}{3}
Mixed number form: x=112,13
x=1\frac{1}{2} , \frac{1}{3}
Decimal form: x=1.5,0.333
x=1.5 , 0.333

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
|5x4|=|x+2|
without the absolute value bars:

|x|=|y||5x4|=|x+2|
x=+y(5x4)=(x+2)
x=y(5x4)=(x+2)
+x=y(5x4)=(x+2)
x=y(5x4)=(x+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||5x4|=|x+2|
x=+y , +x=y(5x4)=(x+2)
x=y , x=y(5x4)=(x+2)

2. Solve the two equations for x

11 additional steps

(5x-4)=(x+2)

Subtract from both sides:

(5x-4)-x=(x+2)-x

Group like terms:

(5x-x)-4=(x+2)-x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

4x4=2

Add to both sides:

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

Simplify the arithmetic:

4x=2+4

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

12 additional steps

(5x-4)=-(x+2)

Expand the parentheses:

(5x-4)=-x-2

Add to both sides:

(5x-4)+x=(-x-2)+x

Group like terms:

(5x+x)-4=(-x-2)+x

Simplify the arithmetic:

6x-4=(-x-2)+x

Group like terms:

6x-4=(-x+x)-2

Simplify the arithmetic:

6x4=2

Add to both sides:

(6x-4)+4=-2+4

Simplify the arithmetic:

6x=2+4

Simplify the arithmetic:

6x=2

Divide both sides by :

(6x)6=26

Simplify the fraction:

x=26

Find the greatest common factor of the numerator and denominator:

x=(1·2)(3·2)

Factor out and cancel the greatest common factor:

x=13

3. List the solutions

x=32,13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x4|
y=|x+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.