Eliminate t using a cancelling linear combination

Dividing the x-equation by 2 gives $x/2=-2t-1/(2t)$. Adding the y-equation gives $x/2+y=t$, so

$t=\frac{x+2y}{2}.$

The original equations require $t\ne0$.

$x=-2(x+2y)-\frac2{x+2y}.$

Multiplying gives $x^2+2xy=-2x^2-8xy-8y^2-2$, hence

$3x^2+10xy+8y^2+2=0.$

The result factors as $(3x+4y)(x+2y)+2=0$, so no real point on it can have $x+2y=0$. Thus the original nonzero-t restriction is automatically enforced; no extra component was introduced when clearing the denominator.