Integer Points Only

Note that $\frac{16x + 80}{2x+11}$ = $8 + \frac{-8}{2x+11}$ .

For $y$ to be an integer, the term $\frac{-8}{2x+11}$ should also be an integer.

That is, $2x+11$ should divide $-8$ .

The factors of $-8$ are $\pm 1, \pm 2, \pm 4$ and $\pm8$ .

However, $2x +11$ can only take the value $\pm 1$ for $x$ to be an integer, .

Thus, there are only $2$ possible values for $x$ and $y$ .