Given y=3x/(x^2-9) determine the numbers m and n if y=m/(x-3)+n/(x+3)

From enunciation, we'll get 2 equivalent expressions for
y:

3x/(x^2 -9)= [m/(x-3)] +
[n/(x+3)]

Since the LCD of the fractions from the right
side is (x-3)(x+3) = x^2 - 9, we'll multiply by x^2 - 9 both
fractions:

3x/(x^2 -9)= [m(x+3) + n(x-3)]/ (x^2
-9)

Having the common denominator (x^2 -9), we'll simplify
it.

3x = mx+3m+nx-3n

We'll
factorize by x to the right side:

3x = x*(m+n) +
(3m-3n)

The terms from the right side and the left side of
the equality, have to be equal so that:

m+n=3
(1)

3m-3n=0

We'll divide by
3:

m - n = 0 (2)

We'll add the
second relation to the first
one:

m+n+m-n=3+0

2m=3

m
= 3/2

But, from (2) => m=n =
3/2

The numbers m and n are equal: m = n =
3/2.