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Solution - Absolute value equations

Exact form: x=2,-16
x=2 , -\frac{1}{6}
Decimal form: x=2,0.167
x=2 , -0.167

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
|8x3|=|4x+5|
without the absolute value bars:

|x|=|y||8x3|=|4x+5|
x=+y(8x3)=(4x+5)
x=y(8x3)=(4x+5)
+x=y(8x3)=(4x+5)
x=y(8x3)=(4x+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||8x3|=|4x+5|
x=+y , +x=y(8x3)=(4x+5)
x=y , x=y(8x3)=(4x+5)

2. Solve the two equations for x

11 additional steps

(8x-3)=(4x+5)

Subtract from both sides:

(8x-3)-4x=(4x+5)-4x

Group like terms:

(8x-4x)-3=(4x+5)-4x

Simplify the arithmetic:

4x-3=(4x+5)-4x

Group like terms:

4x-3=(4x-4x)+5

Simplify the arithmetic:

4x3=5

Add to both sides:

(4x-3)+3=5+3

Simplify the arithmetic:

4x=5+3

Simplify the arithmetic:

4x=8

Divide both sides by :

(4x)4=84

Simplify the fraction:

x=84

Find the greatest common factor of the numerator and denominator:

x=(2·4)(1·4)

Factor out and cancel the greatest common factor:

x=2

12 additional steps

(8x-3)=-(4x+5)

Expand the parentheses:

(8x-3)=-4x-5

Add to both sides:

(8x-3)+4x=(-4x-5)+4x

Group like terms:

(8x+4x)-3=(-4x-5)+4x

Simplify the arithmetic:

12x-3=(-4x-5)+4x

Group like terms:

12x-3=(-4x+4x)-5

Simplify the arithmetic:

12x3=5

Add to both sides:

(12x-3)+3=-5+3

Simplify the arithmetic:

12x=5+3

Simplify the arithmetic:

12x=2

Divide both sides by :

(12x)12=-212

Simplify the fraction:

x=-212

Find the greatest common factor of the numerator and denominator:

x=(-1·2)(6·2)

Factor out and cancel the greatest common factor:

x=-16

3. List the solutions

x=2,-16
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8x3|
y=|4x+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.