\[\large{\begin{cases} x+y+2\sqrt{xy}=64 \\ xy\sqrt{x}+xy\sqrt{y}=32 \end{cases}}\]
Positive real numbers \(x\) and \(y\) satisfy the system of equations above. Find \(x+y\).
Note that we can factor the top expression as
(sqrt (x) + sqrt (y) ) ^2 = 64 take the positive root
(sqrt (x) + sqrt (y) = 8
And the secomd equation cam be factored as
xy ( sqrt (x) + sqrt (y) ) = 32
xy ( 8) = 32
xy = 32/8 = 4
So using the first equarion
x + y + 2sqrt (xy) = 64
x + y + 2sqrt (4) = 64
x + y + 2*2 = 64
x + y + 4 = 64
x + y = 64 - 4 = 60