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

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We are given that y = 3x / (x^2 - 9). We have to determine m and n if y = m/(x - 3) + n/(x +3)

Equate the two expressions for y.

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

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

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

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

equate the coefficients of x and the numeric term

=> 3 = m + n and 3m - 3n = 0

3m - 3n = 0

=> m = n

subtitute in 3 = m + n

=> 3 = 2m

=> m = 3/2

And n = 3/2

**Therefore m = 3/2 and n = 3/2.**

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.**