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Solution - Absolute value equations

Exact form: x=53,-9
x=\frac{5}{3} , -9
Mixed number form: x=123,-9
x=1\frac{2}{3} , -9
Decimal form: x=1.667,9
x=1.667 , -9

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
|5x3|=|4x+12|
without the absolute value bars:

|x|=|y||5x3|=|4x+12|
x=+y(5x3)=(4x+12)
x=y(5x3)=(4x+12)
+x=y(5x3)=(4x+12)
x=y(5x3)=(4x+12)

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||5x3|=|4x+12|
x=+y , +x=y(5x3)=(4x+12)
x=y , x=y(5x3)=(4x+12)

2. Solve the two equations for x

11 additional steps

(5x-3)=(-4x+12)

Add to both sides:

(5x-3)+4x=(-4x+12)+4x

Group like terms:

(5x+4x)-3=(-4x+12)+4x

Simplify the arithmetic:

9x-3=(-4x+12)+4x

Group like terms:

9x-3=(-4x+4x)+12

Simplify the arithmetic:

9x3=12

Add to both sides:

(9x-3)+3=12+3

Simplify the arithmetic:

9x=12+3

Simplify the arithmetic:

9x=15

Divide both sides by :

(9x)9=159

Simplify the fraction:

x=159

Find the greatest common factor of the numerator and denominator:

x=(5·3)(3·3)

Factor out and cancel the greatest common factor:

x=53

8 additional steps

(5x-3)=-(-4x+12)

Expand the parentheses:

(5x-3)=4x-12

Subtract from both sides:

(5x-3)-4x=(4x-12)-4x

Group like terms:

(5x-4x)-3=(4x-12)-4x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x3=12

Add to both sides:

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

Simplify the arithmetic:

x=12+3

Simplify the arithmetic:

x=9

3. List the solutions

x=53,-9
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x3|
y=|4x+12|
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.