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Solution - Absolute value equations

Exact form: x=15,35
x=\frac{1}{5} , \frac{3}{5}
Decimal form: x=0.2,0.6
x=0.2 , 0.6

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
|5x|=|10x+3|
without the absolute value bars:

|x|=|y||5x|=|10x+3|
x=+y(5x)=(10x+3)
x=y(5x)=(10x+3)
+x=y(5x)=(10x+3)
x=y(5x)=(10x+3)

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||5x|=|10x+3|
x=+y , +x=y(5x)=(10x+3)
x=y , x=y(5x)=(10x+3)

2. Solve the two equations for x

7 additional steps

5x=(-10x+3)

Add to both sides:

(5x)+10x=(-10x+3)+10x

Simplify the arithmetic:

15x=(-10x+3)+10x

Group like terms:

15x=(-10x+10x)+3

Simplify the arithmetic:

15x=3

Divide both sides by :

(15x)15=315

Simplify the fraction:

x=315

Find the greatest common factor of the numerator and denominator:

x=(1·3)(5·3)

Factor out and cancel the greatest common factor:

x=15

8 additional steps

5x=-(-10x+3)

Expand the parentheses:

5x=10x3

Subtract from both sides:

(5x)-10x=(10x-3)-10x

Simplify the arithmetic:

-5x=(10x-3)-10x

Group like terms:

-5x=(10x-10x)-3

Simplify the arithmetic:

5x=3

Divide both sides by :

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

Cancel out the negatives:

5x5=-3-5

Simplify the fraction:

x=-3-5

Cancel out the negatives:

x=35

3. List the solutions

x=15,35
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x|
y=|10x+3|
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.