Find real parameter $a$, such that the solution of the linear system lies in the second quadrant 
For which real parameter $a$ lies the solution of the system of equations
$$\begin{aligned} \frac{x}{a+1} + \frac{y}{a-1} &= \frac{1}{a-1}\\ \frac{x}{a+1} - \frac{y}{a-1} &= \frac{1}{a+1} \end{aligned}$$
in the second quadrant?

I do not how to start to solve this system of equations. Any help?

Subtracting the two equations you get
$$\frac{2y}{a-1} = \frac{2}{a^2-1} \Leftrightarrow y = \frac{1}{a+1}, \qquad \text{whenever $a\neq -1$}.$$
Therefore, setting $y$ in the second equation,
$$\frac{x}{a+1} = \frac{1}{a+1}+\frac{y}{a-1} = \frac{1}{a+1} + \frac{1}{a^2-1}=\frac{a}{a^2-1} \Leftrightarrow x = \frac{a}{a-1}, \qquad \text{whenever $a \neq 1$}$$
The solution is actually a point on the cartesian plane. It is the point lying on both lines defining the system of equations. So the question is: for which values of $a$ do we have a negative $x$ coordinate and a positive $y$ coordinate (the conditions that must be met, in order for a point to lie in the second quadrant)?
Well, whenever
$$y \ge 0 \implies a > -1$$
and whenever
$$x \le 0 \Leftrightarrow \frac{a}{a-1}\le 0 \Leftrightarrow 0 \le a \le 1.$$
Because $a=1$ is not allowed (the $x$ coordinate would not be defined) we have that the solution lies in the second quadrant, whenever $0 \le a < 1$.
