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Solution - Absolute value equations

Exact form: x=8,229
x=8 , \frac{22}{9}
Mixed number form: x=8,249
x=8 , 2\frac{4}{9}
Decimal form: x=8,2.444
x=8 , 2.444

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
|3x+1|=|6x23|
without the absolute value bars:

|x|=|y||3x+1|=|6x23|
x=+y(3x+1)=(6x23)
x=y(3x+1)=(6x23)
+x=y(3x+1)=(6x23)
x=y(3x+1)=(6x23)

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||3x+1|=|6x23|
x=+y , +x=y(3x+1)=(6x23)
x=y , x=y(3x+1)=(6x23)

2. Solve the two equations for x

13 additional steps

(3x+1)=(6x-23)

Subtract from both sides:

(3x+1)-6x=(6x-23)-6x

Group like terms:

(3x-6x)+1=(6x-23)-6x

Simplify the arithmetic:

-3x+1=(6x-23)-6x

Group like terms:

-3x+1=(6x-6x)-23

Simplify the arithmetic:

3x+1=23

Subtract from both sides:

(-3x+1)-1=-23-1

Simplify the arithmetic:

3x=231

Simplify the arithmetic:

3x=24

Divide both sides by :

(-3x)-3=-24-3

Cancel out the negatives:

3x3=-24-3

Simplify the fraction:

x=-24-3

Cancel out the negatives:

x=243

Find the greatest common factor of the numerator and denominator:

x=(8·3)(1·3)

Factor out and cancel the greatest common factor:

x=8

10 additional steps

(3x+1)=-(6x-23)

Expand the parentheses:

(3x+1)=-6x+23

Add to both sides:

(3x+1)+6x=(-6x+23)+6x

Group like terms:

(3x+6x)+1=(-6x+23)+6x

Simplify the arithmetic:

9x+1=(-6x+23)+6x

Group like terms:

9x+1=(-6x+6x)+23

Simplify the arithmetic:

9x+1=23

Subtract from both sides:

(9x+1)-1=23-1

Simplify the arithmetic:

9x=231

Simplify the arithmetic:

9x=22

Divide both sides by :

(9x)9=229

Simplify the fraction:

x=229

3. List the solutions

x=8,229
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|3x+1|
y=|6x23|
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.