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Solution - Absolute value equations

Exact form: x=2,43
x=2 , \frac{4}{3}
Mixed number form: x=2,113
x=2 , 1\frac{1}{3}
Decimal form: x=2,1.333
x=2 , 1.333

Other Ways to Solve

Absolute value equations

Step-by-step explanation

1. Rewrite the equation with one absolute value terms on each side

|x+1||2x+3|=0

Add |2x+3| to both sides of the equation:

|x+1||2x+3|+|2x+3|=|2x+3|

Simplify the arithmetic

|x+1|=|2x+3|

2. 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
|x+1|=|2x+3|
without the absolute value bars:

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

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

3. Solve the two equations for x

7 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

x+1=3

Subtract from both sides:

(x+1)-1=3-1

Simplify the arithmetic:

x=31

Simplify the arithmetic:

x=2

12 additional steps

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

Expand the parentheses:

(-x+1)=2x-3

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

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

Group like terms:

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

Simplify the arithmetic:

3x+1=3

Subtract from both sides:

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

Simplify the arithmetic:

3x=31

Simplify the arithmetic:

3x=4

Divide both sides by :

(-3x)-3=-4-3

Cancel out the negatives:

3x3=-4-3

Simplify the fraction:

x=-4-3

Cancel out the negatives:

x=43

4. List the solutions

x=2,43
(2 solution(s))

5. Graph

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