# Thread: weird inequalities problem with reciprocals on both sides (inequalities)

1. ## weird inequalities problem with reciprocals on both sides (inequalities)

This is question number 9 of James Stewart Mathematics for Precalculus 6th Edition

I already know the method where you substitute every single value from S into the equation but I was wondering if there was just a way to simplify it

2. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

$\frac{1}{x} \le \frac{1}{2} \implies \frac{x}{1} \ge \frac{2}{1}$

3. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

What law is that though?

4. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

$\frac{1}{x} \le \frac{1}{2}$

obvious that the inequality is true for all negative values of x

for positive values of x ...

$2x \cdot \frac{1}{x} \le 2x \cdot \frac{1}{2}$

$2 \le x$

5. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

So we multiplied by the denominator multiple thingy, thanks!

6. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

Originally Posted by skeeter
$\frac{1}{x} \le \frac{1}{2} \implies \frac{x}{1} \ge \frac{2}{1}$
That may not be the case. Consider $x<0$.

7. ## Re: weird inequalities problem with reciprocals on both sides (inequalities)

Originally Posted by Plato
That may not be the case. Consider $x<0$.
Right ... corrected in post #4.