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

Sep 22, 2016

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

${x}^{2} + {x}^{2} + 2 x + 1 = 41$

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

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

$\left(x + 5\right) \left(x - 4\right) = 0$

$x = - 5 \mathmr{and} 4$

$\therefore y = - \left(- 5\right) - 1 \text{ AND } y = - 4 - 1$

$y = 4 \text{ AND } y = - 5$

Our solution set is thus $\left\{- 5 , 4\right\}$ and $\left\{4 , - 5\right\}$.

Hopefully this helps!