How do you solve $3 x^{2} - y^{2} = 11$ $3x^2 - y^2 =1 1$ and $x^{2} + 2 y = 2$ $x^2 + 2y = 2$ using substitution?

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Answer 1 · 2016-05-03T01:20:25

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

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

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

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

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

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

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

$y = - 1 and - 5$ $y = -1 and -5$

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

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

$x = 2 or x = 2 \sqrt{3}$ $x = 2 or x = 2sqrt(3)$

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

Hopefully this helps!

How do you solve 3x2−y2=113x^2 - y^2 =1 1 and x2+2y=2x^2 + 2y = 2 using substitution?