Solving $\frac{2}{x} < 3$? How do I solve $\frac{2}{x} < 3$? My book does it this way- $\frac{2}{x} -3< 0$ which implies $\frac{x-(2/3)}{x} < 0$. It then checks signs of intervals using points x=0 and x= 2/3 and finds x value. I tried doing it differently by taking reciprocals getting $\frac{x}{2} > 3$ which implies x>6. Why is my method wrong?
 A: If $x>0$ your method is fine. However, you forgot to take the reciprocal of $3$. So your resulting inequality should read $x/2 >1/3$ and thus $x>2/3$.
Now for $x<0$ suppose we have the following inequality:
$-3<3$ which is true. If we take the reciprocal of both, we get with your method $-1/3>1/3$ which is not true. This all is because the function $1/x$ is not continuous around $0$ and thus you shouldn't use the reciprocal method when the signs of both sides of the equations are not equal.
So, if you still want to use your method and get to the right answer, you should note that for $x<0$, the left-hand side is always negative and the right-hand side is positive. Hence $x<0$ is also a solution.
A: When you take the reciprocals, you're actually multiplying both sides by positive numbers: 
$$x < y \implies x\cdot\frac{1}{|x||y|} < \frac{y}{|x||y|} \implies \operatorname{sgn}x\cdot \frac{1}{|y|} < \operatorname{sgn}y\cdot\frac{1}{|x|}$$
where $\operatorname{sgn}(\cdot)$ stands for $1$ if the argument is positive and $-1$ if negative. If both sides are either positive or negative, you can ignore the absolute values and the sign. But if exactly one of the sides is negative (namely the left side), you get $ -\frac{1}{y} < -\frac{1}{x}$ so $\frac{1}{y} > \frac{1}{x}$.
That's where you got it wrong. You actually supposed the left side will always be positive. If $x<0$, you must reverse the last inequality. Also, you forgot to change $3$ to $\frac{1}{3}$.
A: If $x$ is negative, then the inequalities clearly hold. 
Now for positive $x$, recall that $\frac1x$ is a decreasing function, then by inverting both sides, we have $$\frac{x}2 >\frac13.$$
Hence $x<0$ or $x > \frac23$.
