+1418 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 32 + 2xy$.
For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).
+1725 Squaring the first equation gives [(x+y)^2=x^2+2xy+y^2,]so 102=32+2xy or 2xy=26. Dividing both sides of x+y=10 by 2, we get x/2+y/2=5, so x+y=10. Thus, (x,y) lies on the line y=10−x. Also, since 2xy=26, xy=13, so (x,y) lies on the hyperbola xy=13. The solutions are (x,y)=(2,8), (−2,12), (4,6), (−4,14).
Therefore, the ordered pairs (x,y) are (2,8),(4,6),(8,2),(12,−2),(14,−4)
