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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
|3x+1|=|5x+15|
without the absolute value bars:

|x|=|y||3x+1|=|5x+15|
x=+y(3x+1)=(5x+15)
x=y(3x+1)=(5x+15)
+x=y(3x+1)=(5x+15)
x=y(3x+1)=(5x+15)

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||3x+1|=|5x+15|
x=+y , +x=y(3x+1)=(5x+15)
x=y , x=y(3x+1)=(5x+15)

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:

2x+1=15

Subtract from both sides:

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

Simplify the arithmetic:

2x=151

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

Move the negative sign from the denominator to the numerator:

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:

8x+1=15

Subtract from both sides:

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

Simplify the arithmetic:

8x=151

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=|3x+1|
y=|5x+15|
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.