SOLUTION: I have tried this problem for rational equations and I am unable to get the "work shown" correct. Please help. Thank you Original problem: (N - 1) / (N^2 - 4) = 6 / (N + 2)

Algebra ->  Rational-functions -> SOLUTION: I have tried this problem for rational equations and I am unable to get the "work shown" correct. Please help. Thank you Original problem: (N - 1) / (N^2 - 4) = 6 / (N + 2)      Log On


   



Question 81878: I have tried this problem for rational equations and I am unable to get the "work shown" correct. Please help. Thank you
Original problem: (N - 1) / (N^2 - 4) = 6 / (N + 2)

(n - 1) / (n^2 - 4) = 6 / (n + 2)

denomin: (n^2 - 4), (n + 2)
(n + 2)(n - 2) LCD: (n + 2)(n - 2)

I again need some guidence from here on, I think I get mixed up sometimes with the examples.
multiply each of the two "fractions" by (n+2)(n-2) / (n+2)(n-2) cancelling all that you can first.

CORRECTION:

when I multiply (n+2)(n-2) / (n+2)(n-2), I come up with, 0 (zero),
(n + 2) (n - 2) * (n - 2)(n + 2) cancelling cancels all. I do not think this is correct.

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
you're right about the LCD, multiplying the right hand side by (n-2)/(n-2) will produce the same denominator on both sides

multiplying the equation (by LCD) to clear the denominator leaves (n-1)=6(n-2)

so n-1=6n-12 ... 11=5n ... n=11/5