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Today in class my teacher said that when solving an inequality where the unknown is in the denominator such as: $$\frac {1}{x} < 3$$ You need to follow these steps:

$$x\neq 0 $$

Solve the equation $\frac {1}{x} = 3$, which is $x=\frac{1}{3}$

Then test the the different "regions" on the number line (i.e between 0 and 0.3, greater than 0.3 or less than 0 and see if the inequality holds true).e.g.

Test x = -1, $-1<3$ (Yes)

Test x = 0.25, $4<3$ (No)

Test x = 1, $1<3$ (Yes)

$$\therefore x<0, x>\frac{1}{3}$$

Is there another way to solve this type of inequality without having to test the various values?

Thanks

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  • $\begingroup$ Sketching the graphs of $y=\frac{1}{x}$ and $y=3$ may be useful $\endgroup$ – Kelvin Soh Feb 17 '16 at 6:04
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You can split it into cases, but it's trickier.

Suppose $x>0$. Then $\displaystyle{1\over x}<3$ implies that $1<3x$, or $\displaystyle x>{1\over3}$.

Suppose $x<0$. Then $\displaystyle{1\over x}<3$ implies that $1>3x$ or that $\displaystyle x<{1\over3}$.

Thus, $\displaystyle{1\over x}<3$ when $x>0$ and $\displaystyle x>{1\over3}$, or when $x<0$ and $\displaystyle x<{1\over3}$. This simplifies to $\displaystyle x>{1\over3}$ or $x<0$.

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  • $\begingroup$ Awesome, and this requires no testing? Do you know what this method is called? $\endgroup$ – frog1944 Feb 17 '16 at 6:12
  • $\begingroup$ It gets more difficult if the denominator factors into something like $x(x-1)(x+1)$; you need to consider four cases here ($x<-1$, $-1<x<0$, $0<x<1$, and $x>1$). I don't know that it has a name. $\endgroup$ – Christopher Carl Heckman Feb 18 '16 at 6:11

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