Solve the following equations: $x + y = 7 + \sqrt {xy}$, $x^{2} + y^{2} = 133 - xy$
- $x=9, y=4$
- $x=4, y=8$
- $x=2, y-7$
- $x=5, y=12$
Substitute the given options into the equations. For x=9 and y=4: x+y = 13 and 7+sqrt(36) = 7+6 = 13 (matches). Also, x^2+y^2 = 81+16 = 97, and 133-xy = 133-36 = 97 (matches).
Using the identity (x + y)^2 = x^2 + y^2 + 2xy, we can square the first equation x + y = 7 + sqrt(xy) to get x^2 + y^2 + 2xy = 49 + 14(sqrt(xy)) + xy. From the second equation, we know x^2 + y^2 = 133 - xy, so substituting this gives 133 - xy + 2xy = 49 + 14(sqrt(xy)) + xy. Simplifying this results in 84 = 14(sqrt(xy)), so sqrt(xy) = 6 and xy = 36. Substituting xy = 36 back into the second equation gives x^2 + y^2 = 133 - 36 = 97, and solving the system of x + y = 13 and xy = 36 yields x = 9 and y = 4.