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Solution - Absolute value equations

Exact form: x=52,53
x=\frac{5}{2} , \frac{5}{3}
Mixed number form: x=212,123
x=2\frac{1}{2} , 1\frac{2}{3}
Decimal form: x=2.5,1.667
x=2.5 , 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
|x|=5|x2|
without the absolute value bars:

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

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

2. Solve the two equations for x

11 additional steps

x=5·(x-2)

Expand the parentheses:

x=5x+5·-2

Simplify the arithmetic:

x=5x10

Subtract from both sides:

x-5x=(5x-10)-5x

Simplify the arithmetic:

-4x=(5x-10)-5x

Group like terms:

-4x=(5x-5x)-10

Simplify the arithmetic:

4x=10

Divide both sides by :

(-4x)-4=-10-4

Cancel out the negatives:

4x4=-10-4

Simplify the fraction:

x=-10-4

Cancel out the negatives:

x=104

Find the greatest common factor of the numerator and denominator:

x=(5·2)(2·2)

Factor out and cancel the greatest common factor:

x=52

12 additional steps

x=5·(-(x-2))

Expand the parentheses:

x=5·(-x+2)

x=5·-x+5·2

Group like terms:

x=(5·-1)x+5·2

Multiply the coefficients:

x=-5x+5·2

Simplify the arithmetic:

x=5x+10

Add to both sides:

x+5x=(-5x+10)+5x

Simplify the arithmetic:

6x=(-5x+10)+5x

Group like terms:

6x=(-5x+5x)+10

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=52,53
(2 solution(s))

4. Graph

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