Hello, todd85!

- . . . . . 3y² .= .x . [1]

y² + (x+2)² .= .4 . [2]

Multiply [2] by 3: . 3y² + 3(x+2)² .= .12

. . . Subtract [1]: .-3y² . . . . . . . . = .-x

And we have: . 3(x + 2)² .= .12 - x . → . 3x² + 13x .= .0

Hence: . x(3x + 13) .= .0 . → . x .= .0, -13/3

In [1], we see that x cannot be negative.

Therefore, the only solution is: .x = 0, .y = 0

Since they insist on Elimination, the approach is trickier.

. .xy .= .30 . 1 . . .[1]

x²/6 .= .56 - 2y² . [2]

From [1], we have: .x .= .30/y . → . x² .= .900/y² . [3]

From [2], we have: .x² + 12y² .= .336 . [4]

Subtract [3] from [4]: .12y² .= .336 - 900/y² . → . 12y² - 336 + 900/y² .= .0

Multiply by y²/12: . y^4 - 27y² - 75 .= .0

Factor: . (y² - 3)(y² - 25) .= .0

. - -.- . . . . . . . . _

Hence: . y .= .±√3, .±5

. - - - . . . . . . . . . ._

. .And: . x .= .±10√3, .±6