I really need help solving this system:
\(\begin{align*} x + y &= 8 \\ x^2 + y^2 &= 38 \end{align*}\)
x = 8-y
(8-y)^2 + y^2 = 38
y^2 + 64 + y^2 - 16y = 38
2y^2 - 16y + 26 = 0
y^2 - 8y + 13 = 0 y = \( {8 \pm \sqrt{64-52} \over 2a} = {8 \pm \sqrt{12} \over 2a}\) = (8+-2√3)/2 , which is 4+-√3.
(x, y) = (4-√3, 4+√3), (4+√3, 4-√3)
