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Solution - Absolute value equations

Exact form: x=-12,18
x=-\frac{1}{2} , \frac{1}{8}
Decimal form: x=0.5,0.125
x=-0.5 , 0.125

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
|3x1|=|5x|
without the absolute value bars:

|x|=|y||3x1|=|5x|
x=+y(3x1)=(5x)
x=y(3x1)=(5x)
+x=y(3x1)=(5x)
x=y(3x1)=(5x)

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||3x1|=|5x|
x=+y , +x=y(3x1)=(5x)
x=y , x=y(3x1)=(5x)

2. Solve the two equations for x

10 additional steps

(3x-1)=5x

Subtract from both sides:

(3x-1)-5x=(5x)-5x

Group like terms:

(3x-5x)-1=(5x)-5x

Simplify the arithmetic:

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

Simplify the arithmetic:

2x1=0

Add to both sides:

(-2x-1)+1=0+1

Simplify the arithmetic:

2x=0+1

Simplify the arithmetic:

2x=1

Divide both sides by :

(-2x)-2=1-2

Cancel out the negatives:

2x2=1-2

Simplify the fraction:

x=1-2

Move the negative sign from the denominator to the numerator:

x=-12

7 additional steps

(3x-1)=-5x

Add to both sides:

(3x-1)+1=(-5x)+1

Simplify the arithmetic:

3x=(-5x)+1

Add to both sides:

(3x)+5x=((-5x)+1)+5x

Simplify the arithmetic:

8x=((-5x)+1)+5x

Group like terms:

8x=(-5x+5x)+1

Simplify the arithmetic:

8x=1

Divide both sides by :

(8x)8=18

Simplify the fraction:

x=18

3. List the solutions

x=-12,18
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x1|
y=|5x|
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.