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Solution - Absolute value equations

Exact form: u=-2,23
u=-2 , \frac{2}{3}
Decimal form: u=2,0.667
u=-2 , 0.667

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
|4u|=|2u4|
without the absolute value bars:

|x|=|y||4u|=|2u4|
x=+y(4u)=(2u4)
x=y(4u)=(2u4)
+x=y(4u)=(2u4)
x=y(4u)=(2u4)

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||4u|=|2u4|
x=+y , +x=y(4u)=(2u4)
x=y , x=y(4u)=(2u4)

2. Solve the two equations for u

7 additional steps

4u=(2u-4)

Subtract from both sides:

(4u)-2u=(2u-4)-2u

Simplify the arithmetic:

2u=(2u-4)-2u

Group like terms:

2u=(2u-2u)-4

Simplify the arithmetic:

2u=4

Divide both sides by :

(2u)2=-42

Simplify the fraction:

u=-42

Find the greatest common factor of the numerator and denominator:

u=(-2·2)(1·2)

Factor out and cancel the greatest common factor:

u=2

8 additional steps

4u=-(2u-4)

Expand the parentheses:

4u=2u+4

Add to both sides:

(4u)+2u=(-2u+4)+2u

Simplify the arithmetic:

6u=(-2u+4)+2u

Group like terms:

6u=(-2u+2u)+4

Simplify the arithmetic:

6u=4

Divide both sides by :

(6u)6=46

Simplify the fraction:

u=46

Find the greatest common factor of the numerator and denominator:

u=(2·2)(3·2)

Factor out and cancel the greatest common factor:

u=23

3. List the solutions

u=-2,23
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4u|
y=|2u4|
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.