Given y=3x/(x^2-9) determine the numbers m and n if y=m/(x-3)+n/(x+3)
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:
We'll divide by 3:
m - n = 0 (2)
We'll add the second relation to the first one:
m = 3/2
But, from (2) => m=n = 3/2
The numbers m and n are equal: m = n = 3/2.