Enter an equation or problem
Camera input is not recognized!

Solution - Absolute value equations

Exact form: n=2,0
n=-2 , 0

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
|2n+1|=|3n1|
without the absolute value bars:

|x|=|y||2n+1|=|3n1|
x=+y(2n+1)=(3n1)
x=y(2n+1)=(3n1)
+x=y(2n+1)=(3n1)
x=y(2n+1)=(3n1)

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||2n+1|=|3n1|
x=+y , +x=y(2n+1)=(3n1)
x=y , x=y(2n+1)=(3n1)

2. Solve the two equations for n

7 additional steps

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

Add to both sides:

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

Group like terms:

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

Simplify the arithmetic:

n+1=(-3n-1)+3n

Group like terms:

n+1=(-3n+3n)-1

Simplify the arithmetic:

n+1=1

Subtract from both sides:

(n+1)-1=-1-1

Simplify the arithmetic:

n=11

Simplify the arithmetic:

n=2

9 additional steps

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

Expand the parentheses:

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

Subtract from both sides:

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

Group like terms:

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

Simplify the arithmetic:

-5n+1=(3n+1)-3n

Group like terms:

-5n+1=(3n-3n)+1

Simplify the arithmetic:

5n+1=1

Subtract from both sides:

(-5n+1)-1=1-1

Simplify the arithmetic:

5n=11

Simplify the arithmetic:

5n=0

Divide both sides by the coefficient:

n=0

3. List the solutions

n=2,0
(2 solution(s))

4. Graph

Each line represents the function of one side of the equation:
y=|2n+1|
y=|3n1|
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.