Absolute value, and multiplying by x

Question

How would you solve $$\left|\frac{-2}{x}\right|<1?$$ I did $-1<-2/x<1$ but my teacher always tells me not to multiply by variables, so what do I do?

Answer 1

I would compute with absolute values longer: $$\Bigl\lvert\frac{-2}x\Bigr\rvert=\frac{2}{\lvert x\rvert}<1\iff\lvert x\rvert>2\iff x>2\enspace\text{or}\enspace x<-2. $$

Answer 2

Note $|-\frac {2}{x}| = |\frac{2}{x}|$ so the inequality, equivalently, is $-1<\frac{2}{x}<1$.

We have $1>\frac{2}{x}$ which implies $x>2$ or $x<0$.

We have $-1<\frac{2}{x}$ which implies $x>0$ or $x<-2$.

Combining these yields a general solution $x>2$ or $x<-2$.

Answer 3

Once you remove the absolute values... Do not multiply the inequality by $x$! Why? Because you do not know whether $x$ is positive or negative (it is obvious why $x$ can't be 0).

You can only cross-multiply without changing the signs when $x$ is positive. Assuming that $x$ is a real number, instead of multiplying the inequality by $x$, multiply both sides by $x^2$ instead ($x^2$ will always be positive. Then you have a quadratic inequality: it won't be difficult to solve.

Hope this helped!

Answer 4

This isn't clever or smooth but it is thorough and will always work.

Divide it in cases:

Case 1: $x > 0$

$|\frac {-2}{x}| = \frac {2}{x} <1 \implies 2 < x$

No worry about multiplying by a variable if you know its sign.

Case 2: $x < 0$

$|\frac{-2}{x}| = \frac {2}{-x} < 1 \implies \frac{2}{-x}x > 1x \implies -2 > x$

Case 3: $x = 0$

$|\frac{-2}{0}|$ is impossible.

So $x < -2$ or $x > 2$.