Let √x=c

Let √y=d

The two given equations become:

c^2d+cd^2=14

c^3+d^3=22

If you cube (c+d), you will get an expression purely based on the two above equations:

(c+d)^3 = c^3+d^3+3c^2d+3cd^2

(c+d)^3 = 22 + 14*3

(c+d)^3 = 64

(c+d) = 4

Since the first equation can also be written as cd(c+d) = 14,

cd = 14/4 = 3.5

If you square (c+d) you get:

(c+d)^2 = c^2 + d^2 + 2cd

But c^2 = x and d^2 = y, and we know (c+d) and cd, so therefore we can rearrange to:

(x+y) = (c+d)^2 - 2cd = 16 - 7

(x+y) = 9