Find all pairs of real numbers (x,y)$ such that $x + y = 6$ and $x^2 + y^2 = 28$. If you find more than one pair, then list your pairs in order by increasing $x$ value, separated by commas. For example, to enter the solutions $(2,4)$ and $(-3,9)$, you would enter "(-3,9),(2,4)" (without the quotation marks).
+128631 x + y = 6 ⇒ y = 6 - x (1)
x^2 + y^2 = 28 (2)
Sub (1) into (2) for y
x^2 + (6 - x)^2 = 28 simplify
x^2 + x^2 - 12x + 36 = 28
2x^2- 12x + 8 = 0 divide through by 2
x^2 - 6x + 4 = 0 complete the square on x
x^2 - 6x + 9 = - 4 + 9
(x - 3)^2 = 5 take both roots
x - 3 = ± √5 add 3 to both sides
x = 3 ± √5
Because of the conjugate property of roots....
When x = 3 + √5, y = 3 - √5
When x = 3 - √5 , y = 3 + √5
