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

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

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

We move all terms to the left:

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


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

m∈R



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

m∈R


We add all the numbers together, and all the variables

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

We calculate fractions


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


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

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

We add all the numbers together, and all the variables

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

We multiply all the terms by the denominator


3m0

We add all the numbers together, and all the variables

3m

Back to the equation:

+(3m)



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

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

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 get rid of parentheses

-9m^2+3m=0

a = -9; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·(-9)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$m_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$m_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

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