# Solve {(y^2+x=12),(x^2+y=12):} ?

Oct 30, 2016

((x= 1/2 (1 + 3 sqrt[5]), y= 1/2 (1 - 3 sqrt[5])),(x= 1/2 (1 - 3 sqrt[5]), y= 1/2 (1 + 3 sqrt[5])))

#### Explanation:

With

$\left\{\begin{matrix}{y}^{2} + x = 12 \\ {x}^{2} + y = 12\end{matrix}\right.$

subtracting both sides

${y}^{2} - {x}^{2} + x - y = 0$ or

$\frac{{y}^{2} - {x}^{2}}{y - x} = 1 = x + y$

Now solving

$\left\{\begin{matrix}{y}^{2} + x = 12 \\ x + y = 1\end{matrix}\right.$

we have the solutions

((x= 1/2 (1 + 3 sqrt[5]), y= 1/2 (1 - 3 sqrt[5])),(x= 1/2 (1 - 3 sqrt[5]), y= 1/2 (1 + 3 sqrt[5])))