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Solution - Absolute value equations

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

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
|w+5|=|3w+1|
without the absolute value bars:

|x|=|y||w+5|=|3w+1|
x=+y(w+5)=(3w+1)
x=y(w+5)=(3w+1)
+x=y(w+5)=(3w+1)
x=y(w+5)=(3w+1)

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||w+5|=|3w+1|
x=+y , +x=y(w+5)=(3w+1)
x=y , x=y(w+5)=(3w+1)

2. Solve the two equations for w

13 additional steps

(w+5)=(3w+1)

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-2w+5=(3w+1)-3w

Group like terms:

-2w+5=(3w-3w)+1

Simplify the arithmetic:

2w+5=1

Subtract from both sides:

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

Simplify the arithmetic:

2w=15

Simplify the arithmetic:

2w=4

Divide both sides by :

(-2w)-2=-4-2

Cancel out the negatives:

2w2=-4-2

Simplify the fraction:

w=-4-2

Cancel out the negatives:

w=42

Find the greatest common factor of the numerator and denominator:

w=(2·2)(1·2)

Factor out and cancel the greatest common factor:

w=2

12 additional steps

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

Expand the parentheses:

(w+5)=-3w-1

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

4w+5=(-3w-1)+3w

Group like terms:

4w+5=(-3w+3w)-1

Simplify the arithmetic:

4w+5=1

Subtract from both sides:

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

Simplify the arithmetic:

4w=15

Simplify the arithmetic:

4w=6

Divide both sides by :

(4w)4=-64

Simplify the fraction:

w=-64

Find the greatest common factor of the numerator and denominator:

w=(-3·2)(2·2)

Factor out and cancel the greatest common factor:

w=-32

3. List the solutions

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

4. Graph

Each line represents the function of one side of the equation:
y=|w+5|
y=|3w+1|
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.