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Solution - Absolute value equations

Exact form: w=25,1
w=25 , 1

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
|3w15|=|2w+10|
without the absolute value bars:

|x|=|y||3w15|=|2w+10|
x=+y(3w15)=(2w+10)
x=y(3w15)=(2w+10)
+x=y(3w15)=(2w+10)
x=y(3w15)=(2w+10)

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||3w15|=|2w+10|
x=+y , +x=y(3w15)=(2w+10)
x=y , x=y(3w15)=(2w+10)

2. Solve the two equations for w

7 additional steps

(3w-15)=(2w+10)

Subtract from both sides:

(3w-15)-2w=(2w+10)-2w

Group like terms:

(3w-2w)-15=(2w+10)-2w

Simplify the arithmetic:

w-15=(2w+10)-2w

Group like terms:

w-15=(2w-2w)+10

Simplify the arithmetic:

w15=10

Add to both sides:

(w-15)+15=10+15

Simplify the arithmetic:

w=10+15

Simplify the arithmetic:

w=25

11 additional steps

(3w-15)=-(2w+10)

Expand the parentheses:

(3w-15)=-2w-10

Add to both sides:

(3w-15)+2w=(-2w-10)+2w

Group like terms:

(3w+2w)-15=(-2w-10)+2w

Simplify the arithmetic:

5w-15=(-2w-10)+2w

Group like terms:

5w-15=(-2w+2w)-10

Simplify the arithmetic:

5w15=10

Add to both sides:

(5w-15)+15=-10+15

Simplify the arithmetic:

5w=10+15

Simplify the arithmetic:

5w=5

Divide both sides by :

(5w)5=55

Simplify the fraction:

w=55

Simplify the fraction:

w=1

3. List the solutions

w=25,1
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3w15|
y=|2w+10|
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.