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

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

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

We move all terms to the left:

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


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

h∈R



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

h∈R


We calculate fractions


(3h2/(9(2h-16)*3(2h+4)))+(-9h2/(9(2h-16)*3(2h+4)))=0


We calculate terms in parentheses: +(3h2/(9(2h-16)*3(2h+4))), so:

3h2/(9(2h-16)*3(2h+4))

We multiply all the terms by the denominator


3h2

We add all the numbers together, and all the variables

3h^2

Back to the equation:

+(3h^2)



We calculate terms in parentheses: +(-9h2/(9(2h-16)*3(2h+4))), so:

-9h2/(9(2h-16)*3(2h+4))

We multiply all the terms by the denominator


-9h2

We add all the numbers together, and all the variables

-9h^2

Back to the equation:

+(-9h^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

3h^2-9h^2=0

We add all the numbers together, and all the variables

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