+1348 Find all ordered pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 64 + xy$.
+20 OK so,
lets first rearrange the second equation: x^2-xy+y^2=64.
and we know that x+y=10.
we can rewrite the second equation like this
(x+y)^2-3xy=64
and since x+y=10,
100-3xy=64
which means that xy=12.
NOW
we solve x+y=10 and xy=12.
we can substitute x as 10-y
so we have
y(10-y)=12
10y-y^2=12
and we get the quadratic
y^2-10y+12=0.
this isnt easily factorable so i will use the quadratic formula.
we get that y=5+sqrt 13, 5-sqrt 13.
so that means
(x,y)=(5+sqrt 13, 5-sqrt 13) and (5-sqrt 13, 5+sqrt 13).
:)
