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

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

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

We move all terms to the left:

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


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

m∈R



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

m∈R


We calculate fractions


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


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

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

We multiply all the terms by the denominator


2m2

We add all the numbers together, and all the variables

2m^2

Back to the equation:

+(2m^2)



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

-9m2/(9(2m-16)*2(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

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

We get rid of parentheses

2m^2-9m^2=0

We add all the numbers together, and all the variables

-7m^2=0

a = -7; b = 0; c = 0;
Δ = b2-4ac
Δ = 02-4·(-7)·0
Δ = 0
Delta is equal to zero, so there is only one solution to the equation

                              Stosujemy wzór:
$m=\frac{-b}{2a}=\frac{0}{-14}=0$