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Solution - Absolute value equations

Exact form: x=8,23
x=8 , \frac{2}{3}
Decimal form: x=8,0.667
x=8 , 0.667

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
|4x+1|=|5x7|
without the absolute value bars:

|x|=|y||4x+1|=|5x7|
x=+y(4x+1)=(5x7)
x=y(4x+1)=(5x7)
+x=y(4x+1)=(5x7)
x=y(4x+1)=(5x7)

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||4x+1|=|5x7|
x=+y , +x=y(4x+1)=(5x7)
x=y , x=y(4x+1)=(5x7)

2. Solve the two equations for x

10 additional steps

(4x+1)=(5x-7)

Subtract from both sides:

(4x+1)-5x=(5x-7)-5x

Group like terms:

(4x-5x)+1=(5x-7)-5x

Simplify the arithmetic:

-x+1=(5x-7)-5x

Group like terms:

-x+1=(5x-5x)-7

Simplify the arithmetic:

x+1=7

Subtract from both sides:

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

Simplify the arithmetic:

x=71

Simplify the arithmetic:

x=8

Multiply both sides by :

-x·-1=-8·-1

Remove the one(s):

x=-8·-1

Simplify the arithmetic:

x=8

12 additional steps

(4x+1)=-(5x-7)

Expand the parentheses:

(4x+1)=-5x+7

Add to both sides:

(4x+1)+5x=(-5x+7)+5x

Group like terms:

(4x+5x)+1=(-5x+7)+5x

Simplify the arithmetic:

9x+1=(-5x+7)+5x

Group like terms:

9x+1=(-5x+5x)+7

Simplify the arithmetic:

9x+1=7

Subtract from both sides:

(9x+1)-1=7-1

Simplify the arithmetic:

9x=71

Simplify the arithmetic:

9x=6

Divide both sides by :

(9x)9=69

Simplify the fraction:

x=69

Find the greatest common factor of the numerator and denominator:

x=(2·3)(3·3)

Factor out and cancel the greatest common factor:

x=23

3. List the solutions

x=8,23
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|4x+1|
y=|5x7|
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.