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Solution - Absolute value equations

Exact form: x=4,-427
x=4 , -\frac{4}{27}
Decimal form: x=4,0.148
x=4 , -0.148

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
|15x4|=|12x+8|
without the absolute value bars:

|x|=|y||15x4|=|12x+8|
x=+y(15x4)=(12x+8)
x=y(15x4)=(12x+8)
+x=y(15x4)=(12x+8)
x=y(15x4)=(12x+8)

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||15x4|=|12x+8|
x=+y , +x=y(15x4)=(12x+8)
x=y , x=y(15x4)=(12x+8)

2. Solve the two equations for x

11 additional steps

(15x-4)=(12x+8)

Subtract from both sides:

(15x-4)-12x=(12x+8)-12x

Group like terms:

(15x-12x)-4=(12x+8)-12x

Simplify the arithmetic:

3x-4=(12x+8)-12x

Group like terms:

3x-4=(12x-12x)+8

Simplify the arithmetic:

3x4=8

Add to both sides:

(3x-4)+4=8+4

Simplify the arithmetic:

3x=8+4

Simplify the arithmetic:

3x=12

Divide both sides by :

(3x)3=123

Simplify the fraction:

x=123

Find the greatest common factor of the numerator and denominator:

x=(4·3)(1·3)

Factor out and cancel the greatest common factor:

x=4

10 additional steps

(15x-4)=-(12x+8)

Expand the parentheses:

(15x-4)=-12x-8

Add to both sides:

(15x-4)+12x=(-12x-8)+12x

Group like terms:

(15x+12x)-4=(-12x-8)+12x

Simplify the arithmetic:

27x-4=(-12x-8)+12x

Group like terms:

27x-4=(-12x+12x)-8

Simplify the arithmetic:

27x4=8

Add to both sides:

(27x-4)+4=-8+4

Simplify the arithmetic:

27x=8+4

Simplify the arithmetic:

27x=4

Divide both sides by :

(27x)27=-427

Simplify the fraction:

x=-427

3. List the solutions

x=4,-427
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|15x4|
y=|12x+8|
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.