Answer:
[tex]\frac{3-2n}{n-4}[/tex]
Step-by-step explanation:
We have been given the expression
[tex]\frac{n^2+10n+21}{n^2+3n-28}-\frac{3n}{n-4}[/tex]
Let us write the numerator and denominator of first expression in factored form using AC method.
[tex]n^2+10n+21\\=n^2+7n+3n+21\\n(n+7)+3(n+7)\\(n+7)(n+3)[/tex]
And the factors of denominator is
[tex]n^2+3n-28\\=n^2+7n-4n-28\\n(n+7)-4(n+7)\\(n+7)(n-4)[/tex]
Therefore, the expression becomes
[tex]\frac{(n+7)(n+3)}{(n+7)(n-4)}-\frac{3n}{n-4}\\\\\text{Cancel the common terms}\\\\\frac{n+3}{n-4}-\frac{3n}{n-4}\\\\\text{Denominator of both rational expression is same}\\\text{hence, we can directly add the numerators}\\\\\frac{n+3-3n}{n-4}\\\\\frac{3-2n}{n-4}[/tex]
Thus, the simplified form is [tex]\frac{3-2n}{n-4}[/tex]
