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

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

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

We move all terms to the left:

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


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

t∈R



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

t∈R


We calculate fractions


(3t2/(9(2t-16)*3(2t+4)))+(-9t2/(9(2t-16)*3(2t+4)))=0


We calculate terms in parentheses: +(3t2/(9(2t-16)*3(2t+4))), so:

3t2/(9(2t-16)*3(2t+4))

We multiply all the terms by the denominator


3t2

We add all the numbers together, and all the variables

3t^2

Back to the equation:

+(3t^2)



We calculate terms in parentheses: +(-9t2/(9(2t-16)*3(2t+4))), so:

-9t2/(9(2t-16)*3(2t+4))

We multiply all the terms by the denominator


-9t2

We add all the numbers together, and all the variables

-9t^2

Back to the equation:

+(-9t^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3t^2-9t^2=0

We add all the numbers together, and all the variables

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