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Solution - Absolute value equations

Exact form: u=43,-6
u=\frac{4}{3} , -6
Mixed number form: u=113,-6
u=1\frac{1}{3} , -6
Decimal form: u=1.333,6
u=1.333 , -6

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
|2u+1|=|u+5|
without the absolute value bars:

|x|=|y||2u+1|=|u+5|
x=+y(2u+1)=(u+5)
x=y(2u+1)=(u+5)
+x=y(2u+1)=(u+5)
x=y(2u+1)=(u+5)

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||2u+1|=|u+5|
x=+y , +x=y(2u+1)=(u+5)
x=y , x=y(2u+1)=(u+5)

2. Solve the two equations for u

9 additional steps

(2u+1)=(-u+5)

Add to both sides:

(2u+1)+u=(-u+5)+u

Group like terms:

(2u+u)+1=(-u+5)+u

Simplify the arithmetic:

3u+1=(-u+5)+u

Group like terms:

3u+1=(-u+u)+5

Simplify the arithmetic:

3u+1=5

Subtract from both sides:

(3u+1)-1=5-1

Simplify the arithmetic:

3u=51

Simplify the arithmetic:

3u=4

Divide both sides by :

(3u)3=43

Simplify the fraction:

u=43

8 additional steps

(2u+1)=-(-u+5)

Expand the parentheses:

(2u+1)=u-5

Subtract from both sides:

(2u+1)-u=(u-5)-u

Group like terms:

(2u-u)+1=(u-5)-u

Simplify the arithmetic:

u+1=(u-5)-u

Group like terms:

u+1=(u-u)-5

Simplify the arithmetic:

u+1=5

Subtract from both sides:

(u+1)-1=-5-1

Simplify the arithmetic:

u=51

Simplify the arithmetic:

u=6

3. List the solutions

u=43,-6
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2u+1|
y=|u+5|
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.