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Solution - Absolute value equations

Exact form: x=-12,-14
x=-\frac{1}{2} , -\frac{1}{4}
Decimal form: x=0.5,0.25
x=-0.5 , -0.25

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
|5x+1|=|7x+2|
without the absolute value bars:

|x|=|y||5x+1|=|7x+2|
x=+y(5x+1)=(7x+2)
x=y(5x+1)=(7x+2)
+x=y(5x+1)=(7x+2)
x=y(5x+1)=(7x+2)

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||5x+1|=|7x+2|
x=+y , +x=y(5x+1)=(7x+2)
x=y , x=y(5x+1)=(7x+2)

2. Solve the two equations for x

11 additional steps

(5x+1)=(7x+2)

Subtract from both sides:

(5x+1)-7x=(7x+2)-7x

Group like terms:

(5x-7x)+1=(7x+2)-7x

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

2x+1=2

Subtract from both sides:

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

Simplify the arithmetic:

2x=21

Simplify the arithmetic:

2x=1

Divide both sides by :

(-2x)-2=1-2

Cancel out the negatives:

2x2=1-2

Simplify the fraction:

x=1-2

Move the negative sign from the denominator to the numerator:

x=-12

12 additional steps

(5x+1)=-(7x+2)

Expand the parentheses:

(5x+1)=-7x-2

Add to both sides:

(5x+1)+7x=(-7x-2)+7x

Group like terms:

(5x+7x)+1=(-7x-2)+7x

Simplify the arithmetic:

12x+1=(-7x-2)+7x

Group like terms:

12x+1=(-7x+7x)-2

Simplify the arithmetic:

12x+1=2

Subtract from both sides:

(12x+1)-1=-2-1

Simplify the arithmetic:

12x=21

Simplify the arithmetic:

12x=3

Divide both sides by :

(12x)12=-312

Simplify the fraction:

x=-312

Find the greatest common factor of the numerator and denominator:

x=(-1·3)(4·3)

Factor out and cancel the greatest common factor:

x=-14

3. List the solutions

x=-12,-14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|5x+1|
y=|7x+2|
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.