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Solution - Absolute value equations

Exact form: x=7,1
x=7 , 1

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|=3|x2|
without the absolute value bars:

|x|=|y||2x+1|=3|x2|
x=+y(2x+1)=3(x2)
x=y(2x+1)=3((x2))
+x=y(2x+1)=3(x2)
x=y(2x+1)=3(x2)

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|=3|x2|
x=+y , +x=y(2x+1)=3(x2)
x=y , x=y(2x+1)=3((x2))

2. Solve the two equations for x

12 additional steps

(2x+1)=3·(x-2)

Expand the parentheses:

(2x+1)=3x+3·-2

Simplify the arithmetic:

(2x+1)=3x-6

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+1=6

Subtract from both sides:

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

Simplify the arithmetic:

x=61

Simplify the arithmetic:

x=7

Multiply both sides by :

-x·-1=-7·-1

Remove the one(s):

x=-7·-1

Simplify the arithmetic:

x=7

15 additional steps

(2x+1)=3·(-(x-2))

Expand the parentheses:

(2x+1)=3·(-x+2)

(2x+1)=3·-x+3·2

Group like terms:

(2x+1)=(3·-1)x+3·2

Multiply the coefficients:

(2x+1)=-3x+3·2

Simplify the arithmetic:

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

5x+1=(-3x+6)+3x

Group like terms:

5x+1=(-3x+3x)+6

Simplify the arithmetic:

5x+1=6

Subtract from both sides:

(5x+1)-1=6-1

Simplify the arithmetic:

5x=61

Simplify the arithmetic:

5x=5

Divide both sides by :

(5x)5=55

Simplify the fraction:

x=55

Simplify the fraction:

x=1

3. List the solutions

x=7,1
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2x+1|
y=3|x2|
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.