# How do you solve x^2 - 2y = 1 and x^2 + 5y = 29 using substitution?

Mar 22, 2016

$x = 3$
$y = 4$

#### Explanation:

${x}^{2} - 2 y = 1$${x}^{2} = 1 + 2 y$
${x}^{2} + 5 y = 29$

Then we substitute the ${x}^{2}$ in the second equation with $1 + 2 y$.

$1 + 2 y + 5 y = 29$
$7 y = 28$
$y = 4$

If $y$ is 4 then
${x}^{2} = 1 + 2 \times 4$
${x}^{2} = 9$
$x = \sqrt{9} = 3$