videosfapaturqz

2022-09-26

Solve $\frac{2}{x}<3$ over the reals.
I was using the method of intervals to solve this Question. I seemed to miss $0$ on the number line while using the method of intervals. Why is $0$ included here in the method of intervals?

$\frac{2}{x}-3<0$
or
$\frac{2-3x}{x}<0,$
which gives the answer: $x>\frac{2}{3}$ or $x<0$

demitereur

First suppose $x>0$. Then
$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x>2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x>2/3.$
If $x>0$ and $x>2/3$, then $x>2/3$. Now suppose that $x<0$. Now when we multiply by $x$ the inequality sign flips. Hence
$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}3x<2\phantom{\rule{thickmathspace}{0ex}}⟺\phantom{\rule{thickmathspace}{0ex}}x<2/3.$
If $x<0$ and $x<2/3$, then $x<0$. Thus the solution set is $\left(-\mathrm{\infty },0\right)\cup \left(2/3,\mathrm{\infty }\right)$

Do you have a similar question?