1/9(2n-16)=1/3(2n+4)

Solution for 1/9(2n-16)=1/3(2n+4) equation:

1/9(2n-16)=1/3(2n+4)

We move all terms to the left:

1/9(2n-16)-(1/3(2n+4))=0


Domain of the equation: 9(2n-16)!=0

n∈R



Domain of the equation: 3(2n+4))!=0

n∈R


We calculate fractions


(3n2/(9(2n-16)*3(2n+4)))+(-9n2/(9(2n-16)*3(2n+4)))=0


We calculate terms in parentheses: +(3n2/(9(2n-16)*3(2n+4))), so:

3n2/(9(2n-16)*3(2n+4))

We multiply all the terms by the denominator


3n2

We add all the numbers together, and all the variables

3n^2

Back to the equation:

+(3n^2)



We calculate terms in parentheses: +(-9n2/(9(2n-16)*3(2n+4))), so:

-9n2/(9(2n-16)*3(2n+4))

We multiply all the terms by the denominator


-9n2

We add all the numbers together, and all the variables

-9n^2

Back to the equation:

+(-9n^2)


We add all the numbers together, and all the variables

3n^2+(-9n^2)=0

We get rid of parentheses

3n^2-9n^2=0

We add all the numbers together, and all the variables

-6n^2=0

a = -6; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-6)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$n=\frac{-b}{2a}=\frac{0}{-12}=0$