# How to solve for x,y this system {(x = y-2),(4y+x^2=16):} ?

May 26, 2017

See below.

#### Explanation:

Calling $x , y$ the two numbers such that $x < y$ we have:

$\left\{\begin{matrix}x = y - 2 \\ 4 y + {x}^{2} = 16\end{matrix}\right.$

Substituting the first equation into the second

$4 \left(x + 2\right) + {x}^{2} = 16$ now solving for $x$ we obtain

$x = - 2 \left(1 \pm \sqrt{3}\right)$ so

$y = 2 - 2 \left(1 \pm \sqrt{3}\right)$

If we discard negative solutions, then

$x = 2 \left(\sqrt{3} - 1\right)$ and

$y = 2 \sqrt{3}$