Is it possible for there to be infinitely many solutions in this system of equations?

Question

$$x+(4/3)z=1$$ $$y-(1/6)z=1/2$$ $$z=(2-a)/(-2-a^2)$$ Where a is a constant

I realize that we need all three equations to be equal for there to be infinitely many solutions. The question wants the answer in terms of a. Is it possible?

Answer 1

$x$ and $y$ are easily written in terms of $z$, and if $a$ is constant, $z$ is already given in terms of $a$. So...there is a unique solution for any real $a$.

Specifically: $$x = 1-\tfrac43(2-a)/(-2-a^2)$$ $$y = \tfrac12+\tfrac16(2-a)/(-2-a^2)$$ $$z = (2-a)/(-2-a^2)$$