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

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

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

We move all terms to the left:

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


Domain of the equation: 9(2a+16)!=0

a∈R



Domain of the equation: 3(2a-4))!=0

a∈R


We calculate fractions


(3a2/(9(2a+16)*3(2a-4)))+(-9a2/(9(2a+16)*3(2a-4)))=0


We calculate terms in parentheses: +(3a2/(9(2a+16)*3(2a-4))), so:

3a2/(9(2a+16)*3(2a-4))

We multiply all the terms by the denominator


3a2

We add all the numbers together, and all the variables

3a^2

Back to the equation:

+(3a^2)



We calculate terms in parentheses: +(-9a2/(9(2a+16)*3(2a-4))), so:

-9a2/(9(2a+16)*3(2a-4))

We multiply all the terms by the denominator


-9a2

We add all the numbers together, and all the variables

-9a^2

Back to the equation:

+(-9a^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3a^2-9a^2=0

We add all the numbers together, and all the variables

-6a^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:
$a=\frac{-b}{2a}=\frac{0}{-12}=0$