First, let's rewrite the given equations in a more standard form:
1) $12 + 2y^2 = 1$
2) $2x^2 + 3y^2 = 74$
Now, let's solve the first equation for $y^2$:
$2y^2 = 1 - 12$
$y^2 = \frac{-11}{2}$
Since $y^2$ cannot be negative, there is no solution for this system
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