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) +...
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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.
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