# How do you solve  x^2 + y^2 = 41 and y - x = -1?

Apr 16, 2016

$\left\{\begin{matrix}\left(x y\right) = \left(5 4\right) \\ \left(x y\right) = \left(- 4 - 5\right)\end{matrix}\right.$

#### Explanation:

From the second equation, we can deduce: $y = x - 1$.

Substitute this in the first equation to get:

$41 = {x}^{2} + {\left(x - 1\right)}^{2} = 2 {x}^{2} - 2 x + 1$

Subtract $41$ from both ends to get:

$2 {x}^{2} - 2 x - 40 = 0$

Divide through by $2$ to get:

${x}^{2} - x - 20 = 0$

Note that $5 \cdot 4 = 20$ and $5 - 4 = 1$, hence:

$0 = {x}^{2} - x - 20 = \left(x - 5\right) \left(x + 4\right)$

So $x = 5$ or $x = - 4$

Hence solutions:

$\left\{\begin{matrix}\left(x y\right) = \left(5 4\right) \\ \left(x y\right) = \left(- 4 - 5\right)\end{matrix}\right.$