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

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

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

We move all terms to the left:

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


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

m∈R



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

m∈R


We calculate fractions


(6m2/(9(2m-16)*3(2m+4)))+(-9m2/(9(2m-16)*3(2m+4)))=0


We calculate terms in parentheses: +(6m2/(9(2m-16)*3(2m+4))), so:

6m2/(9(2m-16)*3(2m+4))

We multiply all the terms by the denominator


6m2

We add all the numbers together, and all the variables

6m^2

Back to the equation:

+(6m^2)



We calculate terms in parentheses: +(-9m2/(9(2m-16)*3(2m+4))), so:

-9m2/(9(2m-16)*3(2m+4))

We multiply all the terms by the denominator


-9m2

We add all the numbers together, and all the variables

-9m^2

Back to the equation:

+(-9m^2)


We add all the numbers together, and all the variables

6m^2+(-9m^2)=0

We get rid of parentheses

6m^2-9m^2=0

We add all the numbers together, and all the variables

-3m^2=0

a = -3; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-3)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{-6}=0$