how solve the inequality $−1\leq \leq1$? How can we solve this equation graphically ? why we talk about the two hyperbole $\frac{1}{x}$ and $\frac{-1}{x}$ ?
I divided this equation on two sub-equations $$ xy\leq1$$ and  $$xy\geq-1$$.
For the first sub-equation  I have two cases when x is greater or not than zero. Also for the second I have two other equations depending if $x \geq0$ or not.
That said, I no longer know how to solve my system of 4 sub equations
 A: It can be written compactly as $|xy| \le 1$. The region is symmetrical about $x$-axis and $y$-axis due to the absolute value.
Clearly $x=0$ and $y=0$ are part of the  solution.
If $x \ne 0$, then we have $|y| \le \frac1{|x|}$
In the first quadrant, it is below the line $y=\frac1x$. The other cases can be obtained by using symmetry.
In the fourth quadrant, it is above $y=-\frac1x$.
In the second quadrant, it is below $y=-\frac1x$.
In the third quadrant, it is above $y=\frac1x$.

A: Dividing by cases, we have that

*

*for $x>0$
$$−1\leq \leq1 \iff -\frac1x \le y \le \frac1x$$

*

*for $x<0$ (direction of inequalities flip)

$$−1\leq \leq1 \iff \frac1x \le y \le -\frac1x$$
and for $x=0 \implies xy=0$.
A: The domain of the solutions is delimited by the two hyperbolas $xy=-1$ and $xy=1$ (each with two branches).
These partition the plane in five regions, one being a "cross" in the middle. As $(0,0)$ is a solution, the whole cross, including boundaries, is the solution set. (There cannot be solutions on the "other side" of a boundary.)
