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Solving inequalities with rational expressions.

  1. Sep 22, 2013 #1
    1. The problem statement, all variables and given/known data
    Identify the solution set of the inequality.

    2. Relevant equations
    5x + 1 / x- 1 ≥ 7


    3. The attempt at a solution
    I multiplied both sides by x - 1 which gave me 5x+1 ≥ 7x - 7
    Then I combined like terms together which gave me 8 ≥ 2x
    I divided both sides by 2 which gave me 4 ≥ x.
    Now that's not the answer. The answer is 1 < x ≤ 4.
    Why is that?
     
  2. jcsd
  3. Sep 22, 2013 #2

    Simon Bridge

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    Substitute x=1 into the first relation and see what happens?
    What happens to it when x < 1?
     
  4. Sep 22, 2013 #3

    ehild

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    Hi Viet, welcome to PF.

    Did you mean [itex]5x+\frac{1}{x}-1\ge7[/itex] or [itex]\frac{5x+1}{x-1}\ge7[/itex] ? Do not forget the parentheses!!!
    Multiplying both sides of an inequality with a negative quantity will flip the inequality over. So you keep the inequality when multiplying by x-1 if x>1, and then you get x ≤4.

    Assuming x<1 you get contradiction.

    ehild
     
  5. Sep 22, 2013 #4

    Ray Vickson

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    Your inequality---as written--- is
    [tex] f(x) = 5x + \frac{1}{x} -1 \geq 7 [/tex]
    The 'boundary' points are where f(x) = 7 exactly, so are given by the roots of the quadratic equation ##5x^2+1 = 8x,## or
    [tex]x = \frac{4}{5} \pm \frac{\sqrt{11}}{5} \doteq 0.133675, \:1.463325 [/tex]
    Note that f(x) < 0 for x < 0, so we need x ≥ 0. For ##x \to 0+## or ##x \to +\infty## we have ##f(x) \to + \infty > 7,## so the solution is ##0 < x \leq 0.144675## and ##x \geq 1.463325##.

    All this is under the assumption that what you wrote is what you meant when read using standard rules. If you had meant (5x + 1)/(x-1) ≥ 7 you would have used parentheses.
     
  6. Sep 22, 2013 #5
    [itex]\frac{5x + 1}{x - 1}[/itex] ≥ 7
    That is what I meant. Sorry. My first time using the forums. I didn't know what I was doing. Now I am still confused on how to solve this. As I mentioned before, in my method of solving this I ended up with a one inequality solution: x ≥ 4
    Why is that that's not the answer? The answer is 1 < x ≤ 4. Where did the 1 come from?
     
  7. Sep 22, 2013 #6

    Simon Bridge

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    Please reread post #2 - try the suggestion and you'll see where the 1 came from.
    Please reread post #3 to see the origin of your mistake.

    Your mistake was to fail to take account of the situation where (x-1)≤0.
     
  8. Sep 22, 2013 #7
    I THINK understand now. Thank you.
    So...
    X is not less than 1 because they are the same value making X greater than 1. Is that right?
     
  9. Sep 22, 2013 #8

    ehild

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    When you multiply an inequality with a negative quantity, it flips over.

    See an example:

    -2<1 is true.

    Multiply both sides by -1: you get 2 < -1 which is false. You have to change "<" to ">".

    2>-1.

    If x-1<0 and you multiply with it, the inequality becomes 5x-1≤7(x-1) --> 6≤2x, x>3 , but you assumed that x was less than 1, so you arrived at contradiction. x can not be less than 1 and greater than 3 at the same time.

    X also cannot be 1, as it would mean zero in the denominator. So x>1.

    ehild
     
  10. Oct 1, 2013 #9
    I completely understand now. Thank you very much and sorry for such a late reply. What you did was work with each equation individually. Sorry fro not understanding so fast. My brain takes in things differently than most people. When using math terms, I'm not that well at interpreting it unless it's shown. Thank you.
     
  11. Oct 1, 2013 #10

    ehild

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    I think most people understand Maths through simple examples. When in doubt doing something, try it on simple numerical examples. It helps.


    ehild
     
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