$$15y-11x=8$$
$$15y=8+11x$$
Therefore, (8+11x) must be divisible by 15 which is coefficient of y since 15 is divisible by 5 and 3 the (8+11x) must be divisible by both 5 and 3.
For a number to be divisible by 5, the number must end in either 5 or 0
Therefore, (8+11x) must produce a number that ends in 5 or 0 or both x and y are greater than zero. Both x and y are greater than 3. i will plug in some values of x
If x = 5
$$\left[8+11\left(5\right)\right]=8+55=63$$\
This is not divisible by 5, so $$x\ne5$$
Even though 63 is divisible by 3

If x=12
$$\left[8+\left(11\cdot12\right)\right]=8+132=140$$
This is divisible by 5, but not divisible by 3, so $$x\ne12$$

If x = 17
$$\left[8+\left(x\cdot17\right)\right]=8+187=195$$
This is divisible by both 5 and 3 so, x =17

$$15y-11\left(7\right)=8$$ $$15y-187=8$$ $$15y=8+187$$ $$\frac{15y}{15}=\frac{195}{15}$$ $$y=13$$ $$So,\ y=13,\ x=17$$ $$x+y=17+13=30$$

$$Answer\ is\ Option\ D$$