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Solution - Absolute value equations

Exact form: x=1,-45
x=1 , -\frac{4}{5}
Decimal form: x=1,0.8
x=1 , -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
|8x+1|=|2x+7|
without the absolute value bars:

|x|=|y||8x+1|=|2x+7|
x=+y(8x+1)=(2x+7)
x=y(8x+1)=(2x+7)
+x=y(8x+1)=(2x+7)
x=y(8x+1)=(2x+7)

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||8x+1|=|2x+7|
x=+y , +x=y(8x+1)=(2x+7)
x=y , x=y(8x+1)=(2x+7)

2. Solve the two equations for x

10 additional steps

(8x+1)=(2x+7)

Subtract from both sides:

(8x+1)-2x=(2x+7)-2x

Group like terms:

(8x-2x)+1=(2x+7)-2x

Simplify the arithmetic:

6x+1=(2x+7)-2x

Group like terms:

6x+1=(2x-2x)+7

Simplify the arithmetic:

6x+1=7

Subtract from both sides:

(6x+1)-1=7-1

Simplify the arithmetic:

6x=71

Simplify the arithmetic:

6x=6

Divide both sides by :

(6x)6=66

Simplify the fraction:

x=66

Simplify the fraction:

x=1

12 additional steps

(8x+1)=-(2x+7)

Expand the parentheses:

(8x+1)=-2x-7

Add to both sides:

(8x+1)+2x=(-2x-7)+2x

Group like terms:

(8x+2x)+1=(-2x-7)+2x

Simplify the arithmetic:

10x+1=(-2x-7)+2x

Group like terms:

10x+1=(-2x+2x)-7

Simplify the arithmetic:

10x+1=7

Subtract from both sides:

(10x+1)-1=-7-1

Simplify the arithmetic:

10x=71

Simplify the arithmetic:

10x=8

Divide both sides by :

(10x)10=-810

Simplify the fraction:

x=-810

Find the greatest common factor of the numerator and denominator:

x=(-4·2)(5·2)

Factor out and cancel the greatest common factor:

x=-45

3. List the solutions

x=1,-45
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|8x+1|
y=|2x+7|
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.