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Solution - Absolute value equations

Exact form: x=214,-52
x=\frac{21}{4} , -\frac{5}{2}
Mixed number form: x=514,-212
x=5\frac{1}{4} , -2\frac{1}{2}
Decimal form: x=5.25,2.5
x=5.25 , -2.5

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
|5x3|=|x+18|
without the absolute value bars:

|x|=|y||5x3|=|x+18|
x=+y(5x3)=(x+18)
x=y(5x3)=(x+18)
+x=y(5x3)=(x+18)
x=y(5x3)=(x+18)

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||5x3|=|x+18|
x=+y , +x=y(5x3)=(x+18)
x=y , x=y(5x3)=(x+18)

2. Solve the two equations for x

9 additional steps

(5x-3)=(x+18)

Subtract from both sides:

(5x-3)-x=(x+18)-x

Group like terms:

(5x-x)-3=(x+18)-x

Simplify the arithmetic:

4x-3=(x+18)-x

Group like terms:

4x-3=(x-x)+18

Simplify the arithmetic:

4x3=18

Add to both sides:

(4x-3)+3=18+3

Simplify the arithmetic:

4x=18+3

Simplify the arithmetic:

4x=21

Divide both sides by :

(4x)4=214

Simplify the fraction:

x=214

12 additional steps

(5x-3)=-(x+18)

Expand the parentheses:

(5x-3)=-x-18

Add to both sides:

(5x-3)+x=(-x-18)+x

Group like terms:

(5x+x)-3=(-x-18)+x

Simplify the arithmetic:

6x-3=(-x-18)+x

Group like terms:

6x-3=(-x+x)-18

Simplify the arithmetic:

6x3=18

Add to both sides:

(6x-3)+3=-18+3

Simplify the arithmetic:

6x=18+3

Simplify the arithmetic:

6x=15

Divide both sides by :

(6x)6=-156

Simplify the fraction:

x=-156

Find the greatest common factor of the numerator and denominator:

x=(-5·3)(2·3)

Factor out and cancel the greatest common factor:

x=-52

3. List the solutions

x=214,-52
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x3|
y=|x+18|
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.