Question

$\frac{4n}{6n - 36}+\frac{5n}{n - 1}=\frac{an^{2}+bn}{3(n - 6)(n - 1)}$, $a = $, $b = $

Explanation:

Step1: Simplify the first fraction's denominator

Factor out 6 from $6n - 36$ to get $6(n - 6)$. So the first - fraction is $\frac{4n}{6(n - 6)}=\frac{2n}{3(n - 6)}$.

Step2: Find a common denominator

The common denominator of $\frac{2n}{3(n - 6)}$ and $\frac{5n}{n - 1}$ is $3(n - 6)(n - 1)$.
Rewrite the fractions: $\frac{2n}{3(n - 6)}\times\frac{n - 1}{n - 1}=\frac{2n(n - 1)}{3(n - 6)(n - 1)}$ and $\frac{5n}{n - 1}\times\frac{3(n - 6)}{3(n - 6)}=\frac{15n(n - 6)}{3(n - 6)(n - 1)}$.

Step3: Add the fractions

$\frac{2n(n - 1)+15n(n - 6)}{3(n - 6)(n - 1)}=\frac{2n^{2}-2n + 15n^{2}-90n}{3(n - 6)(n - 1)}=\frac{(2n^{2}+15n^{2})+(-2n-90n)}{3(n - 6)(n - 1)}=\frac{17n^{2}-92n}{3(n - 6)(n - 1)}$.
Since the fraction is $\frac{An^{2}+Bn}{3(n - 6)(n - 1)}$, comparing with $\frac{17n^{2}-92n}{3(n - 6)(n - 1)}$.

Answer:

$A = 17$
$B=-92$