I had to think about this one! There's no indication of what method to use, so I used several. First the answer, so that you know I'm not wasting your time: x=8.534516158 (making sqrt(x)=2.921389422) and y=6.078610578 (making sqrt(y)=2.465483842). This answer is to the best accuracy my calculator could give.
The first job is to eliminate a variable so that we have a single variable equation:
sqrt(9-sqrt(x))=11-x
Still looks kinda horrible, doesn't it?
Take the two sides of the equation as two separate functions:
f(x)=sqrt(9-sqrt(x)); g(x)=11-x
I've assumed that sqrt is the positive square root only. We also know that neither x nor y can be negative, because we can't calculate the real square root of a negative number. We can plot these roughly to give us a picture of the solution, so that we have both functions plotted on the same graph. The shape of g(x) is a straight line from 11 on the vertical axis to 11 on the x axis. We don't need any more more of the graph than this. The shape of f(x) is one half of the parabolic U shape lying on its side with the vertex of the U at (81,0), cutting the vertical axis at (0,3). Where the half-parabola and the straight line intersect is the solution for x. We can see that the intersection point is between x=8 and 9. The easiest way to demonstrate this is work out f(x) and g(x) round these values and compare them. When x=8, f(x)=2.48 and g(x)=3, and when x=9, f(x)=2.45 and g(x)=2. (8,2.48) is lower than (8,3) but (9,2.45) is higher than (9,2), confirming that the graphs intersect between x=8 and 9. We can continue to put values between these limits into the two functions and so get a more and more accurate value for the intersection point. It's a bit laborious, but, depending on the accuracy of our calculator, we can progressively work to the solution for x. When we've gone as far as we want, we take the square root, and then we have enough information to compute y using the simpler of the two expressions, i.e., 9-sqrt(x). We can then calculate the square root of this value and test the original given equations.
There are other ways to proceed but we end up with quartic equations whatever happens, and we have to reject some solutions to these equations because they don't fit our basic requirements or assumptions. Any other solution is not known to me, I'm sorry! Graphically the quartic equations effectively put the other half of the U curve back which we discarded, and the solutions probably include the negative square root that we also discarded.
