How do you solve the system $y^{2} - 2 x^{2} = 6$ $y^2-2x^2=6$ and $y = - 2 x$ $y=-2x$ ?

Question

1382 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-03T03:20:02

The solution set is $(\sqrt{3}, - 2 \sqrt{3})$ $(sqrt(3), -2sqrt(3))$ and $(- \sqrt{3}, 2 \sqrt{3})$ $(-sqrt(3), 2sqrt(3))$

$y = - 2 x \to {(- 2 x)}^{2} - 2 x^{2} = 6$ $y = -2x -> (-2x)^2 - 2x^2 = 6$

$4 x^{2} - 2 x^{2} = 6$ $4x^2 - 2x^2 = 6$

$2 x^{2} = 6$ $2x^2 = 6$

$x^{2} = 3$ $x^2 = 3$

$x = \pm \sqrt{3}$ $x = +-sqrt(3)$

$y = - 2 x$ $y = -2x$

$y = - 2 (\sqrt{3}) AND y = - 2 (- \sqrt{3})$ $y = -2(sqrt(3))" AND "y = -2(-sqrt(3))$

$y = - 2 \sqrt{3} AND y = 2 \sqrt{3}$ $y = -2sqrt(3)" AND "y = 2sqrt(3)$

Hopefully this helps!

How do you solve the system y2−2x2=6y^2-2x^2=6 and y=−2xy=-2x?