x + sqrt(y)=7

sqrt(x) + y =11

sqrt(y) = 7-x

sqrt(x) = 11-y

y = 49 - 14x + x^2

x = 121 - 22y + y^2   substitute this x into eqn immediately above

y = 49 - 14(121 - 22y + y^2)  + (121 - 22y + y^2)^2

y = 49 - 1694 + 308y - 14y^2 + 121^2 + 22^2y^2 + y^4 + 2(-121*22y + 121y^2 - 22y^3)

y = 49 - 1694 + 308y - 14y^2 + 14641 + 484y^2 + y^4 - 5324y + 242y^2 - 44y^3

y = 49 - 1694 + 308y  + 14641    - 5324y 

y^4 - 44y^3 + 712y^2 - 5017y + 12996 = 0    factorable

(y - 9)*(y^3 - 35y^2 + 397y - 1444) = 0

So (y - 9)=0  is one solution, substituting into either equation yields  (4,9) as an integer solution.

At this point, I did use software to see that the cubic has 3 roots, none of which are integers.  But you could use the rational root theorem, and check all the factors of 1444 (which are + or - {1, 2, 4, 19, 38, 76, 361, 722, 1444}) to determine this.