The expression should first be simplified before they are divided
[tex]\begin{gathered} \frac{4}{3n^2\text{ -24n +45 }}\text{ - }\frac{5}{5n^2-25n} \\ =\text{ }\frac{4}{n^2-8n+15}\text{ - }\frac{5}{5n^2-25n}\text{ ( divide by 3 )} \\ =\frac{4}{(n^2-3n)-(5n+15)}\text{ - }\frac{5}{5n\text{ ( n - 5 )}}\text{ ( factorising the denominators )} \\ =\text{ }\frac{4}{n(n-3\text{ ) - 5 (n - 3 )}}\text{ - }\frac{5}{5n\text{ (n-5)}} \\ =\text{ }\frac{4}{(n-3)(n-5)}\text{ - }\frac{1}{n(n-5)}\text{ ( 5 cancels 5 )} \\ =\text{ Now we need to make it into a single expression by finding the L.C.M } \\ =\text{ }\frac{4n\text{ - (n-3)}}{n(n-3)(n-5)}\text{ = }\frac{4n-n\text{ +3}}{n(n-3)(n-5)}\text{ = }\frac{3n\text{ +3}}{n(n-3)(n-5)} \\ =\text{ The simpliest form therefore is }\frac{3n+3}{n(n-3)(n-5)} \\ \end{gathered}[/tex]