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

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

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

We move all terms to the left:

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


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

m∈R



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

m∈R


We add all the numbers together, and all the variables

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

We calculate fractions


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


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

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

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

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

We multiply all the terms by the denominator


-3m2

We add all the numbers together, and all the variables

-3m^2

Back to the equation:

+(-3m^2)


We add all the numbers together, and all the variables

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

We get rid of parentheses

9m^2-3m^2=0

We add all the numbers together, and all the variables

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