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Solution - Absolute value equations

Exact form: x=-127,-45
x=-\frac{12}{7} , -\frac{4}{5}
Mixed number form: x=-157,-45
x=-1\frac{5}{7} , -\frac{4}{5}
Decimal form: x=1.714,0.8
x=-1.714 , -0.8

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
|4x|=|11x+12|
without the absolute value bars:

|x|=|y||4x|=|11x+12|
x=+y(4x)=(11x+12)
x=y(4x)=(11x+12)
+x=y(4x)=(11x+12)
x=y(4x)=(11x+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||4x|=|11x+12|
x=+y , +x=y(4x)=(11x+12)
x=y , x=y(4x)=(11x+12)

2. Solve the two equations for x

7 additional steps

4x=(11x+12)

Subtract from both sides:

(4x)-11x=(11x+12)-11x

Simplify the arithmetic:

-7x=(11x+12)-11x

Group like terms:

-7x=(11x-11x)+12

Simplify the arithmetic:

7x=12

Divide both sides by :

(-7x)-7=12-7

Cancel out the negatives:

7x7=12-7

Simplify the fraction:

x=12-7

Move the negative sign from the denominator to the numerator:

x=-127

8 additional steps

4x=-(11x+12)

Expand the parentheses:

4x=11x12

Add to both sides:

(4x)+11x=(-11x-12)+11x

Simplify the arithmetic:

15x=(-11x-12)+11x

Group like terms:

15x=(-11x+11x)-12

Simplify the arithmetic:

15x=12

Divide both sides by :

(15x)15=-1215

Simplify the fraction:

x=-1215

Find the greatest common factor of the numerator and denominator:

x=(-4·3)(5·3)

Factor out and cancel the greatest common factor:

x=-45

3. List the solutions

x=-127,-45
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x|
y=|11x+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.