Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: w=-32
w=-\frac{3}{2}
Mixed number form: w=-112
w=-1\frac{1}{2}
Decimal form: w=1.5
w=-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
|4w+7|=|4w+5|
without the absolute value bars:

|x|=|y||4w+7|=|4w+5|
x=+y(4w+7)=(4w+5)
x=y(4w+7)=(4w+5)
+x=y(4w+7)=(4w+5)
x=y(4w+7)=(4w+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||4w+7|=|4w+5|
x=+y , +x=y(4w+7)=(4w+5)
x=y , x=y(4w+7)=(4w+5)

2. Solve the two equations for w

5 additional steps

(4w+7)=(4w+5)

Subtract from both sides:

(4w+7)-4w=(4w+5)-4w

Group like terms:

(4w-4w)+7=(4w+5)-4w

Simplify the arithmetic:

7=(4w+5)-4w

Group like terms:

7=(4w-4w)+5

Simplify the arithmetic:

7=5

The statement is false:

7=5

The equation is false so it has no solution.

12 additional steps

(4w+7)=-(4w+5)

Expand the parentheses:

(4w+7)=-4w-5

Add to both sides:

(4w+7)+4w=(-4w-5)+4w

Group like terms:

(4w+4w)+7=(-4w-5)+4w

Simplify the arithmetic:

8w+7=(-4w-5)+4w

Group like terms:

8w+7=(-4w+4w)-5

Simplify the arithmetic:

8w+7=5

Subtract from both sides:

(8w+7)-7=-5-7

Simplify the arithmetic:

8w=57

Simplify the arithmetic:

8w=12

Divide both sides by :

(8w)8=-128

Simplify the fraction:

w=-128

Find the greatest common factor of the numerator and denominator:

w=(-3·4)(2·4)

Factor out and cancel the greatest common factor:

w=-32

3. Graph

Each line represents the function of one side of the equation:
y=|4w+7|
y=|4w+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.