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

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Answer 1 · 2016-03-22T18:39:57

$x = 3$ $x=3$
$y = 4$ $y=4$

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

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

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

If $y$ $y$ is 4 then
$x^{2} = 1 + 2 \times 4$ $x^2=1+2xx4$
$x^{2} = 9$ $x^2=9$
$x = \sqrt{9} = 3$ $x=sqrt9=3$

How do you solve x2−2y=1x^2 - 2y = 1 and x2+5y=29x^2 + 5y = 29 using substitution?