Question

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

Answer 1

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

Explanation:

`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!