1/9(2m-18)=1/3(2m+4)

Solution for 1/9(2m-18)=1/3(2m+4) equation:

1/9(2m-18)=1/3(2m+4)

We move all terms to the left:

1/9(2m-18)-(1/3(2m+4))=0


Domain of the equation: 9(2m-18)!=0

m∈R



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

m∈R


We calculate fractions


(3m2/(9(2m-18)*3(2m+4)))+(-9m2/(9(2m-18)*3(2m+4)))=0


We calculate terms in parentheses: +(3m2/(9(2m-18)*3(2m+4))), so:

3m2/(9(2m-18)*3(2m+4))

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 calculate terms in parentheses: +(-9m2/(9(2m-18)*3(2m+4))), so:

-9m2/(9(2m-18)*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

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

We get rid of parentheses

3m^2-9m^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:
$m=\frac{-b}{2a}=\frac{0}{-12}=0$