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Solution - Absolute value equations

Exact form: x=8,-92
x=8 , -\frac{9}{2}
Mixed number form: x=8,-412
x=8 , -4\frac{1}{2}
Decimal form: x=8,4.5
x=8 , -4.5

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|=|x+17|
without the absolute value bars:

|x|=|y||3x+1|=|x+17|
x=+y(3x+1)=(x+17)
x=y(3x+1)=(x+17)
+x=y(3x+1)=(x+17)
x=y(3x+1)=(x+17)

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|=|x+17|
x=+y , +x=y(3x+1)=(x+17)
x=y , x=y(3x+1)=(x+17)

2. Solve the two equations for x

11 additional steps

(3x+1)=(x+17)

Subtract from both sides:

(3x+1)-x=(x+17)-x

Group like terms:

(3x-x)+1=(x+17)-x

Simplify the arithmetic:

2x+1=(x+17)-x

Group like terms:

2x+1=(x-x)+17

Simplify the arithmetic:

2x+1=17

Subtract from both sides:

(2x+1)-1=17-1

Simplify the arithmetic:

2x=171

Simplify the arithmetic:

2x=16

Divide both sides by :

(2x)2=162

Simplify the fraction:

x=162

Find the greatest common factor of the numerator and denominator:

x=(8·2)(1·2)

Factor out and cancel the greatest common factor:

x=8

12 additional steps

(3x+1)=-(x+17)

Expand the parentheses:

(3x+1)=-x-17

Add to both sides:

(3x+1)+x=(-x-17)+x

Group like terms:

(3x+x)+1=(-x-17)+x

Simplify the arithmetic:

4x+1=(-x-17)+x

Group like terms:

4x+1=(-x+x)-17

Simplify the arithmetic:

4x+1=17

Subtract from both sides:

(4x+1)-1=-17-1

Simplify the arithmetic:

4x=171

Simplify the arithmetic:

4x=18

Divide both sides by :

(4x)4=-184

Simplify the fraction:

x=-184

Find the greatest common factor of the numerator and denominator:

x=(-9·2)(2·2)

Factor out and cancel the greatest common factor:

x=-92

3. List the solutions

x=8,-92
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x+1|
y=|x+17|
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.