Let's suppose that $x, y$ are a positive natural number such that $3x+4y = 95$. How can we maximize and minimize $x+y$? In addition, how many pairs of such $(x, y)$ solving this equation are there?
Writing $4y = 95-3x$,
$$x+y = \frac{4(x+y)}{4} = \frac{4x+4y}{4} = \frac{x+95}{4}$$
Which means that the larger $x$ is chosen, the larger $x+y$ becomes. Then by choosing $y = 2$, we get $x = 29$ and so $x+y = 31$ is maximum. Similarly, choosing $x = 1$ as small as possible, we get $y = 23$ and so $x+y = 24$ is minimum. Does this look reasonable?
All the solutions of $3 x + 4 y = 95$ are $x = -95 + 4 s$ and $y = 95 - 3 s$, where $s\in {\mathbb Z}$, and it follows that $s = x+y$. If $x\ge 0$ and $y\ge 0$, this implies $24\le s\le 31$, so your solution is correct.