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Solution - Absolute value equations

Exact form: x=7,2
x=7 , 2

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|=|5x15|
without the absolute value bars:

|x|=|y||3x1|=|5x15|
x=+y(3x1)=(5x15)
x=y(3x1)=(5x15)
+x=y(3x1)=(5x15)
x=y(3x1)=(5x15)

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|=|5x15|
x=+y , +x=y(3x1)=(5x15)
x=y , x=y(3x1)=(5x15)

2. Solve the two equations for x

13 additional steps

(3x-1)=(5x-15)

Subtract from both sides:

(3x-1)-5x=(5x-15)-5x

Group like terms:

(3x-5x)-1=(5x-15)-5x

Simplify the arithmetic:

-2x-1=(5x-15)-5x

Group like terms:

-2x-1=(5x-5x)-15

Simplify the arithmetic:

2x1=15

Add to both sides:

(-2x-1)+1=-15+1

Simplify the arithmetic:

2x=15+1

Simplify the arithmetic:

2x=14

Divide both sides by :

(-2x)-2=-14-2

Cancel out the negatives:

2x2=-14-2

Simplify the fraction:

x=-14-2

Cancel out the negatives:

x=142

Find the greatest common factor of the numerator and denominator:

x=(7·2)(1·2)

Factor out and cancel the greatest common factor:

x=7

12 additional steps

(3x-1)=-(5x-15)

Expand the parentheses:

(3x-1)=-5x+15

Add to both sides:

(3x-1)+5x=(-5x+15)+5x

Group like terms:

(3x+5x)-1=(-5x+15)+5x

Simplify the arithmetic:

8x-1=(-5x+15)+5x

Group like terms:

8x-1=(-5x+5x)+15

Simplify the arithmetic:

8x1=15

Add to both sides:

(8x-1)+1=15+1

Simplify the arithmetic:

8x=15+1

Simplify the arithmetic:

8x=16

Divide both sides by :

(8x)8=168

Simplify the fraction:

x=168

Find the greatest common factor of the numerator and denominator:

x=(2·8)(1·8)

Factor out and cancel the greatest common factor:

x=2

3. List the solutions

x=7,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x1|
y=|5x15|
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.