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Solution - Absolute value equations

Exact form: x=7,-85
x=7 , -\frac{8}{5}
Mixed number form: x=7,-135
x=7 , -1\frac{3}{5}
Decimal form: x=7,1.6
x=7 , -1.6

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
|6x+1|=|4x+15|
without the absolute value bars:

|x|=|y||6x+1|=|4x+15|
x=+y(6x+1)=(4x+15)
x=y(6x+1)=(4x+15)
+x=y(6x+1)=(4x+15)
x=y(6x+1)=(4x+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||6x+1|=|4x+15|
x=+y , +x=y(6x+1)=(4x+15)
x=y , x=y(6x+1)=(4x+15)

2. Solve the two equations for x

11 additional steps

(6x+1)=(4x+15)

Subtract from both sides:

(6x+1)-4x=(4x+15)-4x

Group like terms:

(6x-4x)+1=(4x+15)-4x

Simplify the arithmetic:

2x+1=(4x+15)-4x

Group like terms:

2x+1=(4x-4x)+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=142

Simplify the fraction:

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

(6x+1)=-(4x+15)

Expand the parentheses:

(6x+1)=-4x-15

Add to both sides:

(6x+1)+4x=(-4x-15)+4x

Group like terms:

(6x+4x)+1=(-4x-15)+4x

Simplify the arithmetic:

10x+1=(-4x-15)+4x

Group like terms:

10x+1=(-4x+4x)-15

Simplify the arithmetic:

10x+1=15

Subtract from both sides:

(10x+1)-1=-15-1

Simplify the arithmetic:

10x=151

Simplify the arithmetic:

10x=16

Divide both sides by :

(10x)10=-1610

Simplify the fraction:

x=-1610

Find the greatest common factor of the numerator and denominator:

x=(-8·2)(5·2)

Factor out and cancel the greatest common factor:

x=-85

3. List the solutions

x=7,-85
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|6x+1|
y=|4x+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.