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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?

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?

Madden Huber

Beginner2022-09-27Added 12 answers

$\frac{2}{x}-3<0$

or

$\frac{2-3x}{x}<0,$

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

or

$\frac{2-3x}{x}<0,$

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

demitereur

Beginner2022-09-28Added 2 answers

First suppose $x>0$. Then

$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x>2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x>2/3.$

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

$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x<2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x<2/3.$

If $x<0$ and $x<2/3$, then $x<0$. Thus the solution set is $(-\mathrm{\infty},0)\cup (2/3,\mathrm{\infty})$

$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x>2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x>2/3.$

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

$\frac{2}{x}<3\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}3x<2\phantom{\rule{thickmathspace}{0ex}}\u27fa\phantom{\rule{thickmathspace}{0ex}}x<2/3.$

If $x<0$ and $x<2/3$, then $x<0$. Thus the solution set is $(-\mathrm{\infty},0)\cup (2/3,\mathrm{\infty})$

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