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Rational inequality question

  1. Oct 9, 2011 #1
    1. The problem statement, all variables and given/known data

    2x-1
    _____ > 0
    5x+3

    2. Relevant equations



    3. The attempt at a solution

    Just wondering, my teacher taught us that youre only supposed to look at what makes the denominator = 0, and dont look at the numerator because it has no affect on anything.

    So, if i were to solve that id have to use -3/5

    so, case 1 would be x < -3/5 which is -0.6, so if i plug in a number less like -1 into the bottom id get a negetive value, however am i plugging it in the top as well...? she said just look @ bot


    or, suppose i have something with two vertical asymptotes like..
    _(x+5)_______ < 0
    (x+3) (x-1)

    Im just looking at the values that make it negetive right? so -3, and 1. if i were to set a interval chart up and say x < -3 , i plug like -4 into (x+3) and into (x-1) and id get a negetive by a negetive which makes it positive. so the sign wouldnt flip on case 1
     
  2. jcsd
  3. Oct 9, 2011 #2

    ehild

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    Gold Member

    Consider the possible signs of both the numerator and the denominator. When is the fraction positive?

    ehild
     
  4. Oct 9, 2011 #3
    This is how

    2x-1
    _____ > 0
    5x+3
    was solved. She said DONT look @ the numerator, just at what the denominator = 0. so

    case 1: x < 2 ( since its negetive at the bottom the sign flips

    so 3x +1 < 0

    x < -1/3

    case 2: x > 2 , sign will pos so no affect on ineq.

    so 3x+1 > 0

    x > -1/3.

    x>2 is the stronger condition so the final statement was -1/3 < x < 2
    So do i only look @ denominator when theres a x^2 on top and bottom to?
     
  5. Oct 9, 2011 #4
    I don't think the advice of "don't look at the numerator" is very useful.

    The expression will change sign when the denominator is zero, but also when the numerator is zero.

    I don't understand all what you're trying to compute. x = 2, and x = -1/3 are NOT values where either the numerator or the denominator changes sign.

    Are you sure your solution is of the same problem as the one in your first post?
     
    Last edited: Oct 9, 2011
  6. Oct 9, 2011 #5
    Oh, musta been a dif example. k , so.

    2x-1
    _____ > 0
    5x+3


    since we only look @ denominator you plug in -3/5 in the bottom and get two cases

    case 1 will be x < -3/5 which is -0.6, so if i plug in -2 . 5(-2) +3 the value is negetive so inequality will flip


    c1 : x < -3/5

    2x-1
    _____ > 0
    5x+3

    2x-1 < 0
    2x < 1

    x < 1/2


    c2: x > -3/5

    2x-1
    _____ > 0
    5x+3

    2x-1 > 0
    x > 1/2

    so i know those are the proper answers, but i dont know how to really set up the therefore statement. Like , if you look at it on a number line all those values seem to intersect at some point

    ie) x < -3/5 , x < 1/2 . this conforms, they go the same direction, but how do i know which one is stronger? Dont we have more information if we know x < 1/2, or do we have more info if we know x < -3/5 ? same for the second case.. How do i figure out what case is predominant?

    -0.6<-----------------0.5 c1 ( on a # line, roughly) (these intersect)

    c2: -0.6 0.5 ----------------> (is 1/2 stronger?)

    so final statement would be -0.6<x<0.5 ?
     
  7. Oct 9, 2011 #6
    This is terriby confusing, and you've managed to flip a sign somewhere, because for x=1, the whole expression is 1/8 wich is bigger than 0, so x=1 should be included in the answer.

    just answer these questions.

    1. when is the denominator >0
    2. when is the numerator >0

    and then

    3. when do the denominator and the numerator have the same sign?
    (there 2 cases here, both positive and both negative)

    draw the intervals on 2 numberlines above each other if you have to.
     
  8. Oct 9, 2011 #7
    Thats the way we learned it, through cases. Yes, when the denom = 0 then inequality sign flips, thats why it flipped. Still right tho?
     
  9. Oct 9, 2011 #8
    the answer you calculated is the solution of

    [tex] \frac {2x-1} { 5x + 3} < 0 [/tex]
     
  10. Oct 9, 2011 #9

    symbolipoint

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    Homework Helper
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    Gold Member

    The basic advice you need is in post #5 of https://www.physicsforums.com/showthread.php?t=538060 . The current inequality,

    [tex]\frac {2x-1} { 5x + 3} < 0[/tex]

    Is simpler and easier to solve. It has maybe two critical values to use. Check all three intervals.
     
  11. Oct 9, 2011 #10

    berkeman

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    Staff: Mentor

    Nelo should be back in about 10 days. Thank you to everybody helping him in this thread.
     
  12. Oct 9, 2011 #11

    HallsofIvy

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    Thanks to you for putting him out of our misery!
     
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