How do you solve a system of $2$ inequalities $xy<1$ and $2+x(y-1)>0$

Question

How do you find all solutions of the following system of inequalities assuming positive and real $x$ and $y$?

$$xy<1$$ $$2 + x(y-1)>0$$

I have rearranged the first inequality to get $y<\frac{1}{x}$ and solved the second inequality for $y$ to get $y>1-\frac{2}{x}$

Thus $\frac{1}{x} \ge 1- \frac{2}{x} \Rightarrow x \le 3$ and the first one gives $y< \frac{1}{3}$.

Similarly solving the second equation for $x$ gives $x> \frac{2}{1-y}$ so assuming that $y \ne 1$ we get $\frac{1}{y} \ge \frac{2}{1-y}$ and then $y \le \frac{1}{3}$ and the first one gives $x<3$.

Now assuming $y=1$, we get $x<1$.

So the solutions found are:

$$x \le 3 \text{ AND } y< \frac{1}{3}$$ $$x<3 \text{ AND } y \le \frac{1}{3}$$ $$x <1 \text{ AND } y=1$$

This seemed like a good method but I have missed the solution $x=1,y<1$

How do you solve such a system to include all solutions? Thanks.

Answer 1

You solve each inequality for $y$: $$\begin{cases}y<\frac1x \\ y>1-\frac{2}{x} \end{cases}$$ Draw feasible region for each inequality and find the overlapping feasible region. Since the feasible region is not rectangular, you can not express the solution numerically. You can write the two simple areas as follows: $$(0<x<2 \ \ \text{AND} \ \ 0<y<\frac1x) \ \ \text{OR} \\ (2\le x< 3 \ \ \text{AND} \ \ 1-\frac2x<y<\frac1x).$$

Answer 2

If you solve both inequalities for $y$, you can think of each inequality as area below (or above) the graph of two functions $y=f(x)$, $y=g(x)$. Since these functions are continuous on $(0,\infty)$, set them equal to each other to find where they intersect. On each interval around the points of intersections, either both inequalities will be true or at least one will be false. Plug in a test point into each function to determine this. Now the inequality will be true on the intervals for which it was true for your test point.