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

Question

1185 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-03T01:20:25

$x^2 = 2 - 2y$

$x = sqrt(2 - 2y)$

$3(sqrt(2 - 2y))^2 - y^2 = 11$

$3(2 - 2y) - y^2 = 11$

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

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

$0 = (y + 1)(y + 5)$

$y = -1 and -5$

$x^2 + 2(-1) = 2 or x^2 + 2(-5) = 2$

$x^2 = 4 or x^2 = 12$

$x = 2 or x = 2sqrt(3)$

Your solution sets are ${2, -1} and {2sqrt(3), -5}$ .

Hopefully this helps!

How do you solve 3x^2 - y^2 =1 13x^2 - y^2 =1 1 and x^2 + 2y = 2x^2 + 2y = 2 using substitution?