Answer:
[tex]\frac{3n^2-7n+15}{(n+3)(n-4)}[/tex] will be the answer.
Step-by-step explanation:
The given expression is,
[tex]\frac{3n}{(n+3)}+\frac{5}{(n-4)}[/tex]
By solving this expression,
[tex]\frac{3n}{(n+3)}+\frac{5}{(n-4)}[/tex]
= [tex]\frac{3n(n-4)}{(n+3)(n-4)}+\frac{5(n+3)}{(n+3)(n-4)}[/tex]
= [tex]\frac{3n(n-4)+5(n+3)}{(n+3)(n-4)}[/tex]
= [tex]\frac{3n^2-12n+5n+15}{(n+3)(n-4)}[/tex]
= [tex]\frac{3n^2-7n+15}{(n+3)(n-4)}[/tex]
Therefore, fraction given in option (2) will be the answer.
