Question: I am stuggling with this inequality and it seems to have tripped up my professor too.
$$\frac{3}{x}-2<8$$

My initial answer was
$$\frac{3}{x}<10$$
$$3<10x$$
$$\frac{3}{10}<x$$
Then I checked the solution sheet my professor provided and he found 
$$x>\frac{3}{10}, x<-\frac{3}{10}, x\ne0$$
Curious, I checked Wolfram Alfpha, MathPapa and Symbolab and they all disagreed 
$$x<0, x>\frac{3}{10}$$
I'm assuming that the websites (computer algebra systems?) are correct and both my professor and I are wrong but I still don't know how to approach such problems systematically. How can I ensure I don't make mistakes like this is the future?

Answer: 
Multiply both sides by $x^2$ to get $3x < 10x^2$. Rearrange and factorize the terms to get $x(10x-3)>0$. Can you continue from here?
$$\frac{3}{x}-2<8\quad\implies 3-2x<8x\quad\implies 3<10x\quad \implies x>\frac{3}{10}<x$$

However multiplying both sides of $3<10x$ by $-1$ reverses things: $$-3>-10x\quad\implies -x<-\frac{3}{10}>-x$$
and here it looks like your professor forgot to change the sign of $x$.

You could square both sides and take the square root of that and get $$x^2=\bigg(-\frac{3}{10}\bigg)^2\quad\implies x<\sqrt{\bigg(-\frac{3}{10}\bigg)^2}=\bigg(\pm\frac{3}{10}\bigg)>x$$

but that's fudging things and does not mean the latter answer is necessarily correct.