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Solution - Absolute value equations

Exact form: x=15,13
x=\frac{1}{5} , \frac{1}{3}
Decimal form: x=0.2,0.333
x=0.2 , 0.333

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
|5x2|=|10x3|
without the absolute value bars:

|x|=|y||5x2|=|10x3|
x=+y(5x2)=(10x3)
x=y(5x2)=(10x3)
+x=y(5x2)=(10x3)
x=y(5x2)=(10x3)

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||5x2|=|10x3|
x=+y , +x=y(5x2)=(10x3)
x=y , x=y(5x2)=(10x3)

2. Solve the two equations for x

11 additional steps

(5x-2)=(10x-3)

Subtract from both sides:

(5x-2)-10x=(10x-3)-10x

Group like terms:

(5x-10x)-2=(10x-3)-10x

Simplify the arithmetic:

-5x-2=(10x-3)-10x

Group like terms:

-5x-2=(10x-10x)-3

Simplify the arithmetic:

5x2=3

Add to both sides:

(-5x-2)+2=-3+2

Simplify the arithmetic:

5x=3+2

Simplify the arithmetic:

5x=1

Divide both sides by :

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

Cancel out the negatives:

5x5=-1-5

Simplify the fraction:

x=-1-5

Cancel out the negatives:

x=15

12 additional steps

(5x-2)=-(10x-3)

Expand the parentheses:

(5x-2)=-10x+3

Add to both sides:

(5x-2)+10x=(-10x+3)+10x

Group like terms:

(5x+10x)-2=(-10x+3)+10x

Simplify the arithmetic:

15x-2=(-10x+3)+10x

Group like terms:

15x-2=(-10x+10x)+3

Simplify the arithmetic:

15x2=3

Add to both sides:

(15x-2)+2=3+2

Simplify the arithmetic:

15x=3+2

Simplify the arithmetic:

15x=5

Divide both sides by :

(15x)15=515

Simplify the fraction:

x=515

Find the greatest common factor of the numerator and denominator:

x=(1·5)(3·5)

Factor out and cancel the greatest common factor:

x=13

3. List the solutions

x=15,13
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x2|
y=|10x3|
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.