# Solve the following system of equations: (x^2+y^2=29),(xy=-10) ?

## Solve the following system of equations: $\left[\begin{matrix}1 \text{ " & x^2+y^2=29 \\ 2 " } & x y = - 10\end{matrix}\right]$?

Jun 10, 2016

#### Answer:

The solutions are $\left\{- 5 , 2\right\} , \left\{- 2 , 5\right\} , \left\{2 , - 5\right\} , \left\{5 , - 2\right\}$

#### Explanation:

Substituting for $y = - \frac{10}{x}$ we have

${x}^{4} - 29 {x}^{2} + 100 = 0$

Making $z = {x}^{2}$ and solving for $z$

${z}^{2} - 29 z + 100 = 0$ and subsequently we have the solutions for $x$

$x = \left\{- 5 , - 2 , 2 , 5\right\}$.

With the final solutions

$\left\{- 5 , 2\right\} , \left\{- 2 , 5\right\} , \left\{2 , - 5\right\} , \left\{5 , - 2\right\}$

The attached figure shows the intersection points of

$\left\{{x}^{2} + {y}^{2} - 20 = 0\right\} \cap \left\{x y + 10 = 0\right\}$