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Solution - Absolute value equations

Exact form: x=12,43
x=\frac{1}{2} , \frac{4}{3}
Mixed number form: x=12,113
x=\frac{1}{2} , 1\frac{1}{3}
Decimal form: x=0.5,1.333
x=0.5 , 1.333

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
|5x5|=|x3|
without the absolute value bars:

|x|=|y||5x5|=|x3|
x=+y(5x5)=(x3)
x=y(5x5)=(x3)
+x=y(5x5)=(x3)
x=y(5x5)=(x3)

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||5x5|=|x3|
x=+y , +x=y(5x5)=(x3)
x=y , x=y(5x5)=(x3)

2. Solve the two equations for x

11 additional steps

(5x-5)=(x-3)

Subtract from both sides:

(5x-5)-x=(x-3)-x

Group like terms:

(5x-x)-5=(x-3)-x

Simplify the arithmetic:

4x-5=(x-3)-x

Group like terms:

4x-5=(x-x)-3

Simplify the arithmetic:

4x5=3

Add to both sides:

(4x-5)+5=-3+5

Simplify the arithmetic:

4x=3+5

Simplify the arithmetic:

4x=2

Divide both sides by :

(4x)4=24

Simplify the fraction:

x=24

Find the greatest common factor of the numerator and denominator:

x=(1·2)(2·2)

Factor out and cancel the greatest common factor:

x=12

12 additional steps

(5x-5)=-(x-3)

Expand the parentheses:

(5x-5)=-x+3

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

6x-5=(-x+3)+x

Group like terms:

6x-5=(-x+x)+3

Simplify the arithmetic:

6x5=3

Add to both sides:

(6x-5)+5=3+5

Simplify the arithmetic:

6x=3+5

Simplify the arithmetic:

6x=8

Divide both sides by :

(6x)6=86

Simplify the fraction:

x=86

Find the greatest common factor of the numerator and denominator:

x=(4·2)(3·2)

Factor out and cancel the greatest common factor:

x=43

3. List the solutions

x=12,43
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x5|
y=|x3|
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.