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Solution - Absolute value equations

Exact form: x=14,12
x=\frac{1}{4} , \frac{1}{2}
Decimal form: x=0.25,0.5
x=0.25 , 0.5

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
|x1|=|5x2|
without the absolute value bars:

|x|=|y||x1|=|5x2|
x=+y(x1)=(5x2)
x=y(x1)=(5x2)
+x=y(x1)=(5x2)
x=y(x1)=(5x2)

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||x1|=|5x2|
x=+y , +x=y(x1)=(5x2)
x=y , x=y(x1)=(5x2)

2. Solve the two equations for x

11 additional steps

(x-1)=(5x-2)

Subtract from both sides:

(x-1)-5x=(5x-2)-5x

Group like terms:

(x-5x)-1=(5x-2)-5x

Simplify the arithmetic:

-4x-1=(5x-2)-5x

Group like terms:

-4x-1=(5x-5x)-2

Simplify the arithmetic:

4x1=2

Add to both sides:

(-4x-1)+1=-2+1

Simplify the arithmetic:

4x=2+1

Simplify the arithmetic:

4x=1

Divide both sides by :

(-4x)-4=-1-4

Cancel out the negatives:

4x4=-1-4

Simplify the fraction:

x=-1-4

Cancel out the negatives:

x=14

12 additional steps

(x-1)=-(5x-2)

Expand the parentheses:

(x-1)=-5x+2

Add to both sides:

(x-1)+5x=(-5x+2)+5x

Group like terms:

(x+5x)-1=(-5x+2)+5x

Simplify the arithmetic:

6x-1=(-5x+2)+5x

Group like terms:

6x-1=(-5x+5x)+2

Simplify the arithmetic:

6x1=2

Add to both sides:

(6x-1)+1=2+1

Simplify the arithmetic:

6x=2+1

Simplify the arithmetic:

6x=3

Divide both sides by :

(6x)6=36

Simplify the fraction:

x=36

Find the greatest common factor of the numerator and denominator:

x=(1·3)(2·3)

Factor out and cancel the greatest common factor:

x=12

3. List the solutions

x=14,12
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|x1|
y=|5x2|
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.