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anonymous

  • one year ago

how do you solve (2x^2-7x+3)/(6x^2+x-2)<0

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  1. Hero
    • one year ago
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    First factor \(2x^2 - 7x + 3\) and \(6x^2 + x - 2\)

  2. anonymous
    • one year ago
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    then what do i do?

  3. Hero
    • one year ago
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    First factor each trinomial. Let me know what you get for each.

  4. anonymous
    • one year ago
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    \[(2x-1)(x-3)/(3x+2)(2x-1)\]

  5. Hero
    • one year ago
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    Next let 1/2, 3, and -3/2 be your critical points.

  6. Hero
    • one year ago
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    Create a number line with those critical points on it. |dw:1438828606993:dw|

  7. anonymous
    • one year ago
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    ok i understand all of that but how am i supposed to answer this?

  8. Hero
    • one year ago
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    Now you have to find the proper intervals that make \[\dfrac{(2x - 1)(x - 3)}{(2x - 1)(3x + 2)} < 0\] true

  9. anonymous
    • one year ago
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    sot th only answer is -3/2 ?

  10. Hero
    • one year ago
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    We haven't gotten to "answers" yet. Hold down ctrl and press the minus key to zoom out.

  11. anonymous
    • one year ago
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    ok

  12. Hero
    • one year ago
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    If you zoom out, you'll see that I created a number line with -3/2, 0, 1/2, and 3 on it.

  13. anonymous
    • one year ago
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    yep i see that

  14. Hero
    • one year ago
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    So, basically, the next step is to pick a number between any of those critical points and "test" it.

  15. anonymous
    • one year ago
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    so btwn -3/2 and 0, i just plug in -1?

  16. Hero
    • one year ago
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    To see if the expression on the left of the inequality is less than zero.

  17. anonymous
    • one year ago
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    ok if i plug -1 in it positive

  18. anonymous
    • one year ago
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    ok if i plug -1 in it positive

  19. Hero
    • one year ago
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    You should start with x = 0. Zero is not a critical point so you can test it.

  20. anonymous
    • one year ago
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    i geta negative number

  21. Hero
    • one year ago
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    And that negative number is obviously less than zero, which means the interval (-3/2, 1/2) is one of the solutions.

  22. anonymous
    • one year ago
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    so do i write the answer like -3/2<x<1/2

  23. Hero
    • one year ago
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    Now understand that the solution will never include two intervals back to back so, try any number greater than 3 to check if 3 < x < ∞ is the other interval.

  24. anonymous
    • one year ago
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    no its positive

  25. Hero
    • one year ago
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    Which number did you test?

  26. anonymous
    • one year ago
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    4

  27. Hero
    • one year ago
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    Test the number 2 and see what you get.

  28. anonymous
    • one year ago
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    i get a negative

  29. Hero
    • one year ago
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    Basically, testing 2 confirms that 1/2 is not a critical point because 2x - 1 is a factor of one.

  30. Hero
    • one year ago
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    So the proper solution interval is actually -3/2 < x < 3

  31. Hero
    • one year ago
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    If you have factors of one in your factorizations, just cancel them out and don't include factors of one in your testing sequence. What I mean by that is, we had \[\dfrac{(2x - 1)(x - 3)}{(2x - 1)(3x + 2)} < 0\] And we should simplify that properly by cancelling out 2x - 1 to get \[\dfrac{x - 3}{3x + 2} < 0\]

  32. Hero
    • one year ago
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    So that's the expression we should have used to test the intervals. And we would only need to use -3/2 and 3 as our critical points.

  33. anonymous
    • one year ago
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    ok i get it! so my final answer would be -3/2<x<3 because btwn those numbers, the equation is less than 0

  34. Hero
    • one year ago
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    The simplified expression (x - 3)/(3x + 2) for the given inequality is less than zero on the interval -3/2 < x < 3. There are no equations here.

  35. anonymous
    • one year ago
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    oh ok ! thanks so much for explaining all of this to me!!!!! you rock!

  36. Hero
    • one year ago
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    You're most welcome.

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