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September 16, 2014

September 16, 2014

Posted by **Clark** on Tuesday, March 30, 2010 at 7:09am.

3n-7/(n-2)(n-3) + 2n+8/9-n^2 - n+2/(n+3)(n-2) = 0

- Math -
**Reiny**, Tuesday, March 30, 2010 at 8:08amWow, I am sure you didn't mean what you typed.

Trusting my instincts, I think you meant

(3n-7)/[(n-2)(n-3)] + (2n+8)/(9-n^2) - (n+2)/[(n+3)(n-2)] = 0

(3n-7)/[(n-2)(n-3)] + 2(n+4)/[(3-n)(3+n)] - (n+2)/[(n+3)(n-2)] = 0

(3n-7)/[(n-2)(n-3)] - 2(n+4)/[(n-3)(3+n)] - (n+2)/[(n+3)(n-2)] = 0

the common denominator is

(n+3)(n-3)(n-2)

so [(3n-7)(n+3) - 2(n+4)(n-2) - (n+2)(n-3)]/[(n+3)(n-3)(n-2)] = 0

(3n-7)(n+3) - 2(n+4)(n-2) - (n+2)(n-3) = 0

Expand then simplify.

You will be amazed at how it breaks apart,

See if you can get n = 1.

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