# How do you solve x^2+2y=22 and x^2-y=13?

Jul 7, 2016

$- y = 13 - {x}^{2}$

$y = - 13 + {x}^{2}$

${x}^{2} + 2 \left(- 13 + {x}^{2}\right) = 22$

${x}^{2} - 26 + 2 {x}^{2} = 22$

$3 {x}^{2} = 22 + 26$

$3 {x}^{2} = 48$

${x}^{2} = 16$

$x = \pm 4$

${4}^{2} - y = 13 \text{ and } - {4}^{2} - y = 13$

$16 - y = 13 \text{ and } 16 - y = 13$

y = 3" and y = 3

The solution set is therefore $\left\{4 , 3\right\}$ and $\left\{- 4 , 3\right\}$.

Hopefully this helps!

Jul 7, 2016

$x = \pm 4$
$y = 3$

#### Explanation:

${x}^{2} + 2 y = 22$
${x}^{2} - y = 13$
$3 y = 9$
$y = 3$
${x}^{2} = 16$
$x = \pm 4$