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?
 A: Adding and subtracting the equations we obtain
$$\frac{2x}{a+1}  = \frac1{a-1}+\frac1{a+1}\implies x=\frac12 \frac{a+1}{a-1}+\frac12=\frac a{a-1}$$
$$\frac{2y}{a-1}  = \frac1{a-1}-\frac1{a+1}\implies y = \frac12-\frac12 \frac{a-1}{a+1}=\frac 1{a+1}$$
then we need $y>0$ and $x<0$.
A: Solve your system : you should find 
$$x=\frac{a}{a-1},\quad y=\frac{1}{a+1}$$
You want $x<0$, so $a$ and $a-1$ must be of opposite signs. So you should have $a>0>a-1$, hence $0<a<1$, and this leads to $y>0$.
The solution is : $(x,y)$ is in the second quadrant if and only if $0<a<1$, or $0\le a<1$ if you look for the closed second quadrant.
A: 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 $0 \le a < 1$.
