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Solution - Absolute value equations

Exact form: s=3,2
s=-3 , -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
|3s+8|=|s4|
without the absolute value bars:

|x|=|y||3s+8|=|s4|
x=+y(3s+8)=(s4)
x=y(3s+8)=(s4)
+x=y(3s+8)=(s4)
x=y(3s+8)=(s4)

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||3s+8|=|s4|
x=+y , +x=y(3s+8)=(s4)
x=y , x=y(3s+8)=(s4)

2. Solve the two equations for s

11 additional steps

(3s+8)=(-s-4)

Add to both sides:

(3s+8)+s=(-s-4)+s

Group like terms:

(3s+s)+8=(-s-4)+s

Simplify the arithmetic:

4s+8=(-s-4)+s

Group like terms:

4s+8=(-s+s)-4

Simplify the arithmetic:

4s+8=-4

Subtract from both sides:

(4s+8)-8=-4-8

Simplify the arithmetic:

4s=-4-8

Simplify the arithmetic:

4s=-12

Divide both sides by :

(4s)4=-124

Simplify the fraction:

s=-124

Find the greatest common factor of the numerator and denominator:

s=(-3·4)(1·4)

Factor out and cancel the greatest common factor:

s=-3

12 additional steps

(3s+8)=-(-s-4)

Expand the parentheses:

(3s+8)=s+4

Subtract from both sides:

(3s+8)-s=(s+4)-s

Group like terms:

(3s-s)+8=(s+4)-s

Simplify the arithmetic:

2s+8=(s+4)-s

Group like terms:

2s+8=(s-s)+4

Simplify the arithmetic:

2s+8=4

Subtract from both sides:

(2s+8)-8=4-8

Simplify the arithmetic:

2s=4-8

Simplify the arithmetic:

2s=-4

Divide both sides by :

(2s)2=-42

Simplify the fraction:

s=-42

Find the greatest common factor of the numerator and denominator:

s=(-2·2)(1·2)

Factor out and cancel the greatest common factor:

s=-2

3. List the solutions

s=3,2
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3s+8|
y=|s4|
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.