simple algebra

by observation one solution is {4,9}

let\, \, \sqrt{x}= t

let\, \, \sqrt{y}= u

solvin simultaneously we have

t + u^2 = 11
u + t^2 =7

we have
u^3 + ut =11 u ---1
u^3 + u^2t^2 = 7 u^2----2

1- 2
ut - u^2t^2 = 11u - 7u^2

t - ut^2 = 11 - 7u

solvin for u we have

u =\frac{t-11}{t^2-7}

substitutin for u we have
t^4 -14^2+t+38=0

we have t= 2 as a real root and three complex roots

for \,\, \, t= 2 \rightarrow x = 4

for \, t= 2 \rightarrow y = 11-t =11-2 = 9

complex roots u may see wolfram

btw hats off qwerty[123] !!!!!!!!!!!

http://www.goiit.com/posts/list/algebra-open-challenge-to-all-community-members-and-experts-73653.htm

prophet sir is this the only way to do tis?

I really dont know. but this looks simple enuf.

you are right, they will. But the problem as I see it appear time and again usually asks for real soutions only.

setting x =m² and y =n²
m + n² =7 → (1)
m² + n= 11 → (2)

subtracting (2) from (1)

m² + n - m - n² =4

(m-n)(m+n-1) = 4

this eq. has integral sol. only for m=3 nd n=2....... so x = 9 nd y= 4