Find all pairs $(x,y)$ of real numbers such that $x + y = 10$ and $x^2 + y^2 = 56$. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(2,4),(-3,9)" (without the quotation marks).
+128460 x + y = 10 ⇒ y = 10 - x (1)
x^2 + y^2 = 56 (2) sub (1) into (2)
x^2 + (10 - x)^2 = 56
x^2 + x^2 - 20x + 100 = 56
2x^2 - 20x + 44 = 0 divide through by 2
x^2 - 10x + 22 = 0 complete the square on x
x^2 - 10x + 25 = -22 + 25
(x - 5)^2 = 3 take both roots
x - 5 = ± √3
x = ± √3 + 5
Using ( 1) the solutions are ( √3 + 5 , - √3 + 5) and ( - √3 + 5 , √3 + 5)
