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Math Help - inequalities help!!

  1. #1
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    inequalities help!!

    hey guys,


    (3x+5)/(2x-1) (greater than or equal to) 8

    any ideas? thanks guys
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  2. #2
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    Quote Originally Posted by jvignacio View Post
    hey guys,


    (3x+5)/(2x-1) (greater than or equal to) 8

    any ideas? thanks guys
    Note that \frac{3x+5}{2x-1} = \frac{13/2}{2x-1} + \frac{3}{2} = \frac{1}{2} \left( \frac{13}{2x-1} + 3\right).

    Substitute this expression and solve the inequality for x.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Note that \frac{3x+5}{2x-1} = \frac{13/2}{2x-1} + \frac{3}{2} = \frac{1}{2} \left( \frac{13}{2x-1} + 3\right).

    Substitute this expression and solve the inequality for x.
    ok i have just solved it but not sure if its correct. is the answer

    all x's in the intervals (1, negative infinity) and (1/2, positive infinity) ???
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  4. #4
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    Quote Originally Posted by jvignacio View Post
    ok i have just solved it but not sure if its correct. is the answer

    all x's in the intervals (1, negative infinity) and (1/2, positive infinity) ???
    What is (1, negative infinity) meant to mean?

    And by (1/2, positive infinity) do you mean 1/2 < x < +oo ?

    Have you taken some random values of x from your solution and checked to see if they satisfy the original inequality? I have. I took x = 2 from (1/2, positive infinity) (assuming it means what I think it means - see above) and it fails the test. So your solution cannot be correct.

    If you show all of your working the mistakes you've made in getting this solution can be pointed out.

    By the way - don't bump.
    Last edited by mr fantastic; October 6th 2008 at 04:42 AM.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    What is (1, negative infinity) meant to mean?

    And by (1/2, positive infinity) do you mean 1/2 < x < +oo ?

    Have you taken some random values of x from your solution and checked to see if they satisfy the original inequality? I have. I took x = 1 from (1/2, positive infinity) (assuming it means what I think it means - see above) and it fails the test. So your solution cannot be correct.

    If you show all of your working the mistakes you've made in getting this solution can be pointed out.

    By the way - don't bump.
    whats bump?

    ok this is what i did...

    <br />
\frac{3x+5}{2x-1}\geq 8<br />


    <br />
\frac{3x+5}{2x-1}-8\geq 0<br />

    <br />
\frac{3x+5}{2x-1}-\frac{8(2x-1)}{2x-1}\geq 0<br />

    <br />
\frac{3x+5 - 8(2x-1)}{2x-1}\geq 0<br />

    <br />
\frac{-13x+13}{2x-1}\geq 0<br />

    then i got -13(1) + 13 = 0 so 1 and every other negative number below 1 for x will satisfy its equal to or greater than 0.

    and for 2x-1, i got 2(1/2)-1 = 0 so 1/2 and every other positive number above 1/2 will satisfy its equal to or greater than 0.
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  6. #6
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    Quote Originally Posted by jvignacio View Post
    whats bump?

    ok this is what i did...

    <br />
\frac{3x+5}{2x-1}\geq 8<br />


    <br />
\frac{3x+5}{2x-1}-8\geq 0<br />

    <br />
\frac{3x+5}{2x-1}-\frac{8(2x-1)}{2x-1}\geq 0<br />

    <br />
\frac{3x+5 - 8(2x-1)}{2x-1}\geq 0<br />

    <br />
\frac{-13x+13}{2x-1}\geq 0<br />

    [snip]
    Case 1: -13 x + 13 \geq 0 and 2x - 1 > 0, that is, x \leq 1 and x > \frac{1}{2}, that is, \frac{1}{2} < x \leq 1.

    Case 2: -13 x + 13 \leq 0 and 2x - 1 < 0, that is, x \geq 1 and x < \frac{1}{2}, which has no solution.

    So the solution is \frac{1}{2} < x \leq 1.


    By the way: http://en.wikipedia.org/wiki/Bump_(Internet)
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  7. #7
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    Quote Originally Posted by mr fantastic View Post
    Case 1: -13 x + 13 \geq 0 and 2x - 1 > 0, that is, x \leq 1 and x > \frac{1}{2}, that is, \frac{1}{2} < x \leq 1.

    Case 2: -13 x + 13 \leq 0 and 2x - 1 < 0, that is, x \geq 1 and x < \frac{1}{2}, which has no solution.

    So the solution is \frac{1}{2} < x \leq 1.


    By the way: Bump (Internet) - Wikipedia, the free encyclopedia
    i get it! thank u
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