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Solution - Absolute value equations

Exact form: x=-113,97
x=-\frac{11}{3} , \frac{9}{7}
Mixed number form: x=-323,127
x=-3\frac{2}{3} , 1\frac{2}{7}
Decimal form: x=3.667,1.286
x=-3.667 , 1.286

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
|10x+2|=|4x20|
without the absolute value bars:

|x|=|y||10x+2|=|4x20|
x=+y(10x+2)=(4x20)
x=y(10x+2)=(4x20)
+x=y(10x+2)=(4x20)
x=y(10x+2)=(4x20)

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||10x+2|=|4x20|
x=+y , +x=y(10x+2)=(4x20)
x=y , x=y(10x+2)=(4x20)

2. Solve the two equations for x

11 additional steps

(10x+2)=(4x-20)

Subtract from both sides:

(10x+2)-4x=(4x-20)-4x

Group like terms:

(10x-4x)+2=(4x-20)-4x

Simplify the arithmetic:

6x+2=(4x-20)-4x

Group like terms:

6x+2=(4x-4x)-20

Simplify the arithmetic:

6x+2=20

Subtract from both sides:

(6x+2)-2=-20-2

Simplify the arithmetic:

6x=202

Simplify the arithmetic:

6x=22

Divide both sides by :

(6x)6=-226

Simplify the fraction:

x=-226

Find the greatest common factor of the numerator and denominator:

x=(-11·2)(3·2)

Factor out and cancel the greatest common factor:

x=-113

12 additional steps

(10x+2)=-(4x-20)

Expand the parentheses:

(10x+2)=-4x+20

Add to both sides:

(10x+2)+4x=(-4x+20)+4x

Group like terms:

(10x+4x)+2=(-4x+20)+4x

Simplify the arithmetic:

14x+2=(-4x+20)+4x

Group like terms:

14x+2=(-4x+4x)+20

Simplify the arithmetic:

14x+2=20

Subtract from both sides:

(14x+2)-2=20-2

Simplify the arithmetic:

14x=202

Simplify the arithmetic:

14x=18

Divide both sides by :

(14x)14=1814

Simplify the fraction:

x=1814

Find the greatest common factor of the numerator and denominator:

x=(9·2)(7·2)

Factor out and cancel the greatest common factor:

x=97

3. List the solutions

x=-113,97
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|10x+2|
y=|4x20|
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.