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Solution - Absolute value equations

Exact form: x=1,14
x=1 , \frac{1}{4}
Decimal form: x=1,0.25
x=1 , 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
|2x+1|=|6x3|
without the absolute value bars:

|x|=|y||2x+1|=|6x3|
x=+y(2x+1)=(6x3)
x=y(2x+1)=(6x3)
+x=y(2x+1)=(6x3)
x=y(2x+1)=(6x3)

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||2x+1|=|6x3|
x=+y , +x=y(2x+1)=(6x3)
x=y , x=y(2x+1)=(6x3)

2. Solve the two equations for x

12 additional steps

(2x+1)=(6x-3)

Subtract from both sides:

(2x+1)-6x=(6x-3)-6x

Group like terms:

(2x-6x)+1=(6x-3)-6x

Simplify the arithmetic:

-4x+1=(6x-3)-6x

Group like terms:

-4x+1=(6x-6x)-3

Simplify the arithmetic:

4x+1=3

Subtract from both sides:

(-4x+1)-1=-3-1

Simplify the arithmetic:

4x=31

Simplify the arithmetic:

4x=4

Divide both sides by :

(-4x)-4=-4-4

Cancel out the negatives:

4x4=-4-4

Simplify the fraction:

x=-4-4

Cancel out the negatives:

x=44

Simplify the fraction:

x=1

12 additional steps

(2x+1)=-(6x-3)

Expand the parentheses:

(2x+1)=-6x+3

Add to both sides:

(2x+1)+6x=(-6x+3)+6x

Group like terms:

(2x+6x)+1=(-6x+3)+6x

Simplify the arithmetic:

8x+1=(-6x+3)+6x

Group like terms:

8x+1=(-6x+6x)+3

Simplify the arithmetic:

8x+1=3

Subtract from both sides:

(8x+1)-1=3-1

Simplify the arithmetic:

8x=31

Simplify the arithmetic:

8x=2

Divide both sides by :

(8x)8=28

Simplify the fraction:

x=28

Find the greatest common factor of the numerator and denominator:

x=(1·2)(4·2)

Factor out and cancel the greatest common factor:

x=14

3. List the solutions

x=1,14
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x+1|
y=|6x3|
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.