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Math Help - Rational inequalities

  1. #1
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    Rational inequalities

     \frac{1}{2x-3} > 5


    Can someone show me how the heck i work these out, i don't recall going over them in class and i have a test tonight which i need to know how to do this. I know what the answer is but i am banging my head as to how the heck it is that. I know how to do inequalities but not rational ones.
    Last edited by mr fantastic; August 20th 2009 at 03:57 AM.
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  2. #2
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    Quote Originally Posted by el123 View Post
     \frac{1}{2x-3} > 5


    Can someone show me how the heck i work these out, i don't recall going over them in class and i have a test tonight which i need to know how to do this. I know what the answer is but i am banging my head as to how the heck it is that. I know how to do inequalities but not rational ones.
    In order to be greater than 5, \frac{1}{2x-3} must be certainly be greater than 0 so 2x-3 is positive. That means that we can just multiply both sides of the inequality by it to get 1> 5(2x- 3)= 10x- 15. That should be easy to solve.

    That was easy because it was clear that the denominator must e positive to satisfy the inequality but usually we cannot be sure of that. More generally, the simplest way to solve complicated inequalities is to first solve the associated equality- here, \frac{1}{2x-3}= 5. Solving that is also easy: multiplying both sides by 2x- 3, 1= 5(2x- 3). 10x- 15= 1 gives 10x= 16 so x= 16/10= 8/5.

    The point of that is that such an inequality can "swap" directions only where the two sides are equal or where there is a "discontinuity". Rational functions are discontinuous only where the denominator is 0:here where 2x-3= 0 or x= 3/2. The two points at which the inquality can reverse are 3/2 and 8/5. They divide the real number line into three intervals. Checking a single point in each interval will tell you which is ">" and which is "<".
    Last edited by mr fantastic; August 20th 2009 at 03:57 AM. Reason: Edited quote.
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  3. #3
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    the denominator should be  |2x-3|

    Sorry , i tried to edit but net cut out.

    so does the aboe still apply if the denominator is an absolute value?
    Last edited by mr fantastic; August 20th 2009 at 03:58 AM. Reason: Merged posts
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  4. #4
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    Quote Originally Posted by el123 View Post
    so does the aboe still apply if the denominator is an absolute value?
    Hello you have two formula
    The firste is :

    a > 0,b > 0:a > b \Leftrightarrow \frac{1}{a} < \frac{1}{b}<br />
    The second is :

    c > 0:\left| x \right| < c \Leftrightarrow - c < x < c
    In this exercice :

    \frac{1}{{\left| {2x - 3} \right|}} > 5 \Leftrightarrow \left| {2x - 3} \right| < \frac{1}{5}<br />
    \left| {2x - 3} \right| < \frac{1}{5} \Leftrightarrow - \frac{1}{5} < 2x - 3 < \frac{1}{5}

     - \frac{1}{5} < 2x - 3 < \frac{1}{5} \Leftrightarrow - \frac{1}{5} + 3 < 2x - 3 + 3 < \frac{1}{5} + 3
     <br />
- \frac{1}{5} + 3 < 2x - 3 + 3 < \frac{1}{5} + 3 \Leftrightarrow \frac{{14}}{5} < 2x < \frac{{16}}{5}<br /> <br />
    \frac{{14}}{5} < 2x < \frac{{16}}{5} \Leftrightarrow \frac{7}{5} < x < \frac{8}{5}
    Conclusion the set of solutions is :

    \frac{7}{5} < x < \frac{8}{5}
    Last edited by mr fantastic; August 20th 2009 at 03:58 AM. Reason: Fixed close quote tag
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  5. #5
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    Hello, el123!

    \frac{1}{2x-3} \:>\: 5

    Resist the temptation to eliminate the denominator ... It contains an x.

    Instead, bring all terms to one side and simplify:

    . . \frac{1}{2x-3} - 5 \:>\:0 \quad\Rightarrow\quad \frac{16-10x}{2x-3} \:>\:0

    We have a positive fraction.


    A fraction ia positive if:

    . . (1) the numerator and denominator are positive.
    . . (2) the numerator and denominator are negative.


    Examine the two cases.

    (1) Both positive: . \begin{array}{ccccccc}16-10x \:>\:0 & \Rightarrow & x \:<\:\frac{8}{5} \\ \\[-4mm] 2x-3 \:>\:0 & \Rightarrow & x \:>\:\frac{3}{2} \end{array}

    . . .We have: . (x < 1.6) \wedge (x > 1.5) \quad\Rightarrow\quad{\color{blue} \boxed{1.5 \:<\:x \:<\:1.6}}


    (2) Both negative: . \begin{array}{ccccccc}16-10x \:<\:0 & \Rightarrow & x \:>\:\frac{8}{5} \\ \\[-4mm] 2x-3 \:<\:0 & \Rightarrow & x \:<\:\frac{3}{2} \end{array}

    . . .We have: . (x > 1.6) \wedge (x < 1.5)\quad\hdots \text{Impossible}

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