How do you solve the system #y^2-2x^2=6# and #y=-2x#?

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Answer 1 · 2016-10-03T03:20:02

The solution set is #(sqrt(3), -2sqrt(3))# and #(-sqrt(3), 2sqrt(3))#

#y = -2x -> (-2x)^2 - 2x^2 = 6#

#4x^2 - 2x^2 = 6#

#2x^2 = 6#

#x^2 = 3#

#x = +-sqrt(3)#

#y = -2x#

#y = -2(sqrt(3))" AND "y = -2(-sqrt(3))#

#y = -2sqrt(3)" AND "y = 2sqrt(3)#

Hopefully this helps!