>In other words, no matter the input, there is only one solution to the system of equations for that input.

You're almost correct. But if you thought that, why would you say for part (A) that the only values of a which produce exactly one solution for x and y would be a=1 or a=-1. As you said above, almost any value of a will produce a solution.

In fact, if you solve for x and y you should get x=2/a and y=-1/a. So as long as a doesn't equal 0, there's always at least one solution.

But it's possible that you could get more than one solution. Part (A) wants you to give the values of a where x=2/a and y=-1/a are the only solution, and part (B) wants you to give the values of a where there are more than one solution.

If you've used matrices to solve systems of equations, then you'd approach this by writing the system as M[x,y]^(T) = [2a+3,1/a]^(T) where M is the matrix [[a^(2),-3a][1,1]]. Then the system will have exactly one solution unless det(M)=0. det(M) = a^(2)+3a which equals 0 when a=0 or a=-3.

That tells you that the system may have zero or infinitely many solutions when a=-3 (a=0 was already ruled out by the statement of the question).

When a=-3 the system becomes: 9x+9y=-3 and x+y=-1/3 which has infinitely many solutions. Just let x be anything and let y = -1/3-x and (x,y) will satisfy both equations.

Any other value of a will produce exactly 1 solution.

Note: If you haven't used matrices then you'd just solve the system for x and y using other methods (such as subsitution) and you'd get to a point where you'd have to handle the case where a=-3 separately.