Let $a, b, c$ be real numbers. I must consider the following system of linear equations:
$$\begin{cases}x+2by=a\\bx+(1-b)y=c^2+c\end{cases}$$
For which set of values of $a$ is the following proposition true: "For every $b$ there exists a $c$ such that the system has a solution"?
I really have no clue on where to start with this question. Any advice is appreciated.
For a $c$ with a solution to exist, it must be possible satisfy the second equation, which is a quadratic for $c$.
So we know that there is a solution $c$ precisely when the associated discriminant is nonnegative. This discriminant is $\Delta = 1-4(bx+(1-b)y)$.
When we solve for $x$ using the other equation, i.e. $x = a-2by$, we obtain:
\begin{align} \Delta &= 1-4(b(a-2by)+(1-b)y) \\ &= 1-4ab + 8b^2y +y - by\\ &= 1-4ab+y(8b^2-b+1) \end{align}
Because we may vary $y$ to obtain a solution, the only way $\Delta$ is guaranteed to be negative is when $8b^2-b+1$ is zero. This you can use to provide conditions on $a$ to ensure that $\Delta \ge 0$ even in that case.
Bottom line: Establish conditions for the variables whose existence you must show (in this case, $c,x,y$), and deal with the generic (in this case, if $8b^2-b+1 \ne 0$) and special/particular ($8b^2-b+1 = 0$) cases for those variables which can take arbitrary values (here, $b$). The last kind of "variable" is the constant: it's given to you beforehand; $a$ is an example of such a "variable" in this case.