+2669 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 64 + xy$.
+129850 x + y = 10 square both sides
x^2 + 2xy + y^2 = 100
x^2 + y^2 = 100 - 2xy
64 + xy = 100 - 2xy
3xy = 36
xy = 12
y = 12 / x
x + 12 / x = 10
x^2 + 12 = 10x
x^2 - 10x = -12
x^2 -10x + 25 = -12 + 25
(x - 5)^2 = 13 take both roots
x - 5 = sqrt (13) x - 5 = -sqrt 13
x = 5 + sqrt (13) x = 5 - sqrt (13)
y = 12 / ( 5 + sqrt 13) y = 12 / (5 -sqrt 13)
y = [12 (5 -sqrt 13)] / 12 y= 12 (5 + sqrt 13) / ( 12)
y = 5 -sqrt (13) y = 5 + sqrt 13
(x, y) = ( 5 + sqrt 13 , 5 -sqrt 13 ) and ( 5 -sqrt 13 , 5 + sqrt 13)
