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

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    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
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    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}
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    Re: weird inequalities problem with reciprocals on both sides (inequalities)

    What law is that though?
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    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
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    Re: weird inequalities problem with reciprocals on both sides (inequalities)

    So we multiplied by the denominator multiple thingy, thanks!
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    Re: weird inequalities problem with reciprocals on both sides (inequalities)

    Quote Originally Posted by skeeter View Post
    \frac{1}{x} \le \frac{1}{2} \implies \frac{x}{1} \ge \frac{2}{1}
    That may not be the case. Consider $x<0$.
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    Re: weird inequalities problem with reciprocals on both sides (inequalities)

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