Math
posted by Clark .
Solve:
3n7/(n2)(n3) + 2n+8/9n^2  n+2/(n+3)(n2) = 0

Wow, I am sure you didn't mean what you typed.
Trusting my instincts, I think you meant
(3n7)/[(n2)(n3)] + (2n+8)/(9n^2)  (n+2)/[(n+3)(n2)] = 0
(3n7)/[(n2)(n3)] + 2(n+4)/[(3n)(3+n)]  (n+2)/[(n+3)(n2)] = 0
(3n7)/[(n2)(n3)]  2(n+4)/[(n3)(3+n)]  (n+2)/[(n+3)(n2)] = 0
the common denominator is
(n+3)(n3)(n2)
so [(3n7)(n+3)  2(n+4)(n2)  (n+2)(n3)]/[(n+3)(n3)(n2)] = 0
(3n7)(n+3)  2(n+4)(n2)  (n+2)(n3) = 0
Expand then simplify.
You will be amazed at how it breaks apart,
See if you can get n = 1.