Question

(7/(n^2-n-12))-(2/(n^2-6n+8)) I need to simply this as much as possible, can anyone lend me some assistance please?

Answer 1

factor first denominator one as (n+3)(n-4) nd second as (n-2)(n-4) so the lcm of denominator ll be (n+3)(n-4)(n-2), now solve

Answer 2

n^2-n-12= n^2-4n+3n-12=n(n-4)+3(n-4)=(n+3)(n-4)
n^2-6n+8=n^2-4n-2n+8=n(n-4)-2(n-4)=(n-2)(n-4)
7/(n+3)(n-4)-2/(n-2)(n-4)
=1/(n-4)x{7/(n+3)-2(n-2)}
=1/(n-4)x{(7(n-2)-2(n+3))/(n+3)(n-2)}
=1/(n-4)x{(7n-14-2n-6))/(n+3)(n-2)}
=(5n-20)/(n-2)(n+3)(n-4) this is simplification