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

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

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

We move all terms to the left:

1/9(-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


(3m2/(9(-2m-16)*3(2m+4)))+(-9m0/(9(-2m-16)*3(2m+4)))=0


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

3m2/(9(-2m-16)*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: +(-9m0/(9(-2m-16)*3(2m+4))), so:

-9m0/(9(-2m-16)*3(2m+4))

We multiply all the terms by the denominator


-9m0

We add all the numbers together, and all the variables

-9m

Back to the equation:

+(-9m)


We get rid of parentheses

3m^2-9m=0

a = 3; b = -9; c = 0;
Δ = b2-4ac
Δ = -92-4·3·0
Δ = 81
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{81}=9$
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-9}{2*3}=\frac{0}{6}
=0
$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+9}{2*3}=\frac{18}{6}
=3
$