+729 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 64 + xy$.
+129771 x + y = 10 square both sides
x^2 + 2xy + y^2 = 100
x^2 + y^2 = 100 - 2xy
So
100 - 2xy = 64 + xy
100 - 64 = 3xy
36 = 3xy
12 = xy → y = 12/x
So
x + 12/x = 10
x^2 -10x + 12 = 0
x = [ 10 +/- sqrt [100 - 48] ] / [ 2 = [ 10 +/- sqrt [ 52] ]/ 2 = [10 +/- 2sqrt13 ] / 2 = 5 +/- sqrt 13
Using the conjugate property
(x,y) = ( 5 + sqrt 13 , 5 -sqrt 13 ) , (5 -sqrt 13 , 5 + sqrt 13)
