Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: x=-2,213
x=-2 , \frac{2}{13}
Decimal form: x=2,0.154
x=-2 , 0.154

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
|5x4|=|8x+2|
without the absolute value bars:

|x|=|y||5x4|=|8x+2|
x=+y(5x4)=(8x+2)
x=y(5x4)=(8x+2)
+x=y(5x4)=(8x+2)
x=y(5x4)=(8x+2)

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||5x4|=|8x+2|
x=+y , +x=y(5x4)=(8x+2)
x=y , x=y(5x4)=(8x+2)

2. Solve the two equations for x

13 additional steps

(5x-4)=(8x+2)

Subtract from both sides:

(5x-4)-8x=(8x+2)-8x

Group like terms:

(5x-8x)-4=(8x+2)-8x

Simplify the arithmetic:

-3x-4=(8x+2)-8x

Group like terms:

-3x-4=(8x-8x)+2

Simplify the arithmetic:

3x4=2

Add to both sides:

(-3x-4)+4=2+4

Simplify the arithmetic:

3x=2+4

Simplify the arithmetic:

3x=6

Divide both sides by :

(-3x)-3=6-3

Cancel out the negatives:

3x3=6-3

Simplify the fraction:

x=6-3

Move the negative sign from the denominator to the numerator:

x=-63

Find the greatest common factor of the numerator and denominator:

x=(-2·3)(1·3)

Factor out and cancel the greatest common factor:

x=2

10 additional steps

(5x-4)=-(8x+2)

Expand the parentheses:

(5x-4)=-8x-2

Add to both sides:

(5x-4)+8x=(-8x-2)+8x

Group like terms:

(5x+8x)-4=(-8x-2)+8x

Simplify the arithmetic:

13x-4=(-8x-2)+8x

Group like terms:

13x-4=(-8x+8x)-2

Simplify the arithmetic:

13x4=2

Add to both sides:

(13x-4)+4=-2+4

Simplify the arithmetic:

13x=2+4

Simplify the arithmetic:

13x=2

Divide both sides by :

(13x)13=213

Simplify the fraction:

x=213

3. List the solutions

x=-2,213
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x4|
y=|8x+2|
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.