# How do you solve 3x^2 - y^2 =1 1 and x^2 + 2y = 2 using substitution?

May 3, 2016

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

$x = \sqrt{2 - 2 y}$

$3 {\left(\sqrt{2 - 2 y}\right)}^{2} - {y}^{2} = 11$

$3 \left(2 - 2 y\right) - {y}^{2} = 11$

$6 - 6 y - {y}^{2} = 11$

$0 = {y}^{2} + 6 y + 5$

$0 = \left(y + 1\right) \left(y + 5\right)$

$y = - 1 \mathmr{and} - 5$

${x}^{2} + 2 \left(- 1\right) = 2 \mathmr{and} {x}^{2} + 2 \left(- 5\right) = 2$

${x}^{2} = 4 \mathmr{and} {x}^{2} = 12$

$x = 2 \mathmr{and} x = 2 \sqrt{3}$

Your solution sets are $\left\{2 , - 1\right\} \mathmr{and} \left\{2 \sqrt{3} , - 5\right\}$.

Hopefully this helps!