$\frac{dx}{dt} = x-xy,\quad \frac{dy}{dt} = -y+xy.$ Find the range of values of $\ x\ $ and $\ y\ $ for which both variables are increasing. 
$$\frac{dx}{dt} = x-xy,\quad \frac{dy}{dt} = -y+xy.$$
Find the range of values of $\ x\ $ and $\ y\ $ for which both
variables are increasing.

Here, increasing means the derivative is $\ >0.$
My attempt:
$x-xy>0,\quad -y+xy>0 \implies x>y.$
We also have:
$\ x(1-y)>0\ \implies \text{either}\ (\ x>0\ \text{and}\ y<1\ )\quad \text{or}\ (\ x<0\ \text{and}\ y>1\ ).$
and
$y(-1+x)>0\implies\ (\ y>0\ \text{and}\ x>1\ )\quad \text{or}\ (\ y<0\ \text{and}\ x<1\ ),$
But now I don't know where to go from here without getting confused. We can also say that
$\ x(-1+x)>y(-1+x)>0,\ $ assuming that $\ ( \ y>0\ $ and $\ x>1),\ $ or $\ ( \ y<0\ $ and $\ x<1).$
and try to make sense of the quadratic inequality in $x,$ but the different cases make this a headache.
Note that this is a difficult A Level problem, so should be solvable quite quickly. I'm confident that I'm missing something straightforward.
 A: We have $x\gt y$, $xy\gt y$. This implies three cases:

*

*$x,y\lt0,\,x\gt y\implies|x|\lt|y|$. This works because $xy$ is positive, but $y$ is negative.

*$x,y\gt0,\,x\gt1,x\gt y$. This is self-evident.

*$y\lt0, 1\gt x\ge0$. This works because those ranges of $x$ shrink the absolute value of $xy$, making $xy$ greater than $y$ since they are both negative.

But, I neglected to consider $x\gt xy$ there - this narrows the cases as follows:

*

*Case $1$ is now impossible! This is because $x$ is negative, $xy$ is positive.

*Case $2$ has the added condition $y\lt1$.

*Case $3$ needs only $1\gt x\ge0$ changed to $1\gt x\gt0$, now making $x$ strictly positive, since this makes $xy$ negative and $x$ positive.

So the ranges are:
$$\begin{align}&1)\quad x\in(1,\infty),\,y\in(0,1)\\&2)\quad x\in(0,1),\,y\in(-\infty,0)\end{align}$$
This region plot will show this visually:

To test my answer's validity, I had Desmos plot for me in the red zones: $x-xy\gt0$, and in the blue zones: $-y+xy\gt0$, and you are interested in their intersection. The overlapping regions match what I said.
