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Solution - Absolute value equations

Exact form: s=-83,-45
s=-\frac{8}{3} , -\frac{4}{5}
Mixed number form: s=-223,-45
s=-2\frac{2}{3} , -\frac{4}{5}
Decimal form: s=2.667,0.8
s=-2.667 , -0.8

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
|4s+6|=|s2|
without the absolute value bars:

|x|=|y||4s+6|=|s2|
x=+y(4s+6)=(s2)
x=y(4s+6)=(s2)
+x=y(4s+6)=(s2)
x=y(4s+6)=(s2)

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||4s+6|=|s2|
x=+y , +x=y(4s+6)=(s2)
x=y , x=y(4s+6)=(s2)

2. Solve the two equations for s

9 additional steps

(4s+6)=(s-2)

Subtract from both sides:

(4s+6)-s=(s-2)-s

Group like terms:

(4s-s)+6=(s-2)-s

Simplify the arithmetic:

3s+6=(s-2)-s

Group like terms:

3s+6=(s-s)-2

Simplify the arithmetic:

3s+6=-2

Subtract from both sides:

(3s+6)-6=-2-6

Simplify the arithmetic:

3s=-2-6

Simplify the arithmetic:

3s=-8

Divide both sides by :

(3s)3=-83

Simplify the fraction:

s=-83

10 additional steps

(4s+6)=-(s-2)

Expand the parentheses:

(4s+6)=-s+2

Add to both sides:

(4s+6)+s=(-s+2)+s

Group like terms:

(4s+s)+6=(-s+2)+s

Simplify the arithmetic:

5s+6=(-s+2)+s

Group like terms:

5s+6=(-s+s)+2

Simplify the arithmetic:

5s+6=2

Subtract from both sides:

(5s+6)-6=2-6

Simplify the arithmetic:

5s=2-6

Simplify the arithmetic:

5s=-4

Divide both sides by :

(5s)5=-45

Simplify the fraction:

s=-45

3. List the solutions

s=-83,-45
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4s+6|
y=|s2|
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.