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Solution - Absolute value equations

Exact form: u=-32,2
u=-\frac{3}{2} , 2
Mixed number form: u=-112,2
u=-1\frac{1}{2} , 2
Decimal form: u=1.5,2
u=-1.5 , 2

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
|3u1|=|u+5|
without the absolute value bars:

|x|=|y||3u1|=|u+5|
x=+y(3u1)=(u+5)
x=y(3u1)=(u+5)
+x=y(3u1)=(u+5)
x=y(3u1)=(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||3u1|=|u+5|
x=+y , +x=y(3u1)=(u+5)
x=y , x=y(3u1)=(u+5)

2. Solve the two equations for u

13 additional steps

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-4u-1=(u+5)-u

Group like terms:

-4u-1=(u-u)+5

Simplify the arithmetic:

4u1=5

Add to both sides:

(-4u-1)+1=5+1

Simplify the arithmetic:

4u=5+1

Simplify the arithmetic:

4u=6

Divide both sides by :

(-4u)-4=6-4

Cancel out the negatives:

4u4=6-4

Simplify the fraction:

u=6-4

Move the negative sign from the denominator to the numerator:

u=-64

Find the greatest common factor of the numerator and denominator:

u=(-3·2)(2·2)

Factor out and cancel the greatest common factor:

u=-32

14 additional steps

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

Expand the parentheses:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2u1=5

Add to both sides:

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

Simplify the arithmetic:

2u=5+1

Simplify the arithmetic:

2u=4

Divide both sides by :

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

Cancel out the negatives:

2u2=-4-2

Simplify the fraction:

u=-4-2

Cancel out the negatives:

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

3. List the solutions

u=-32,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3u1|
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.