by HallsofIvy » Wed Mar 13, 2019 7:46 am
Another way: The inequality is $\frac{1}{x}+ 3< 0$. Subtracting 3 from both sides, $\frac{1}{x}< -3$.
Now, an important way solving inequalities differs from solving equalities: Multiplying both sides of an inequality by a negative number reverses the direction of the inequality. Multiplying both sides of an inequality by a positive number does not change the direction and, of course, x cannot be 0 here.
So we consider two possibilities. If x> 0, then -3x< 0 so multiplying both sides of $\frac{1}{x}< -3$ by -3x gives $-3> x$ which is the same as $x< -3$. But then x is not positive so there is no solution here. If x< 0 then -3x> 0 so multiplying both sides of $\frac{1}{x}< -3$ bu -3x gives $-3< x$. Since we must have x< 0 for this solution so we must have $-3< x< 0$.
The solution to the inequality is "all x such that -3< x< 0".