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Inequalities with two different fractions which include x in the denominator.

  1. Jan 14, 2013 #1
    1. The problem statement, all variables and given/known data

    (x-2)/(x+3) less than (x+1)/(x)

    3. The attempt at a solution

    I broke it up into cases.

    When x+3 less than 0 and x less than 0

    and then when both are positive, when one is positive and the other is negative and then the other way around.

    I'm not sure if that's the right way though. I was thinking that maybe you can say:

    (x+3)(x) less than 0

    and a second case

    (x+3)(x) greater than 0
     
  2. jcsd
  3. Jan 14, 2013 #2
    (x+3) and x are not independent. You have to consider the value ranges of x.

    Perhaps a good place to start is to consider the value of each expression as they pass the "difficult" value of x in each case, and how those behave as x is large-negative and large-positive in each case again.

    There are three relevant ranges of x, one of which is a little trickier than the other two.
     
  4. Jan 14, 2013 #3
    So you mean to say x≠-3 and x≠0 and x≠ some other number?
     
  5. Jan 14, 2013 #4
    One of the ranges is x < -3
     
  6. Jan 14, 2013 #5
    You mean x < -3 is not in the 'solution set' right? x is -4 makes it not a true statement.

    After trying the first method again (the textbook's method except they didn't show any examples with two fractions both with x in the denominator) I found that x > -3 , x cannot equal -1/2 , 0
     
  7. Jan 14, 2013 #6
    x can equal anything. Of the two expressions being considered in the question, each one is undefined at one value of x (which you already have indicated you know).

    {x<-3} is a range of values where both expressions are well-defined. You might like to explore this range with some values of x (like -4, -5, -10) to see how the functions are behaving.
     
  8. Jan 14, 2013 #7
    When you input -4 for x, you get 6 < 0.75 which means that x < -3 is not a part of the solution right?

    When you input -1/2 for x, you get -1 < -1 which again is not true.

    Although plugging values is good to check answers, I wanted to know if my method was correct. I don't have the answer to this question though.

    What I did was solve for x under all circumstances. When x+3 > 0 and x > 0 and then when x+3 < 0 and x < 0 and then when x+3 > 0 but x < 0 then when x+3 < 0 and x > 0

    The idea behind this is that when solving for x, multiplying by -(x+3) or -(x) would switch the inequality sign. So the possibility that one or the other or both are positive or negative becomes an issue. Is this a correct way to solve this inequality?
     
  9. Jan 14, 2013 #8

    SammyS

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    It seems a lot easier to write [itex]\displaystyle \ \ \frac{x-2}{x+3}<\frac{x+1}{x}[/itex]

    as the equivalent inequality: [itex]\displaystyle \ \ \frac{x-2}{x+3}-\frac{x+1}{x}<0\ .[/itex]

    Then use a common denominator to combine the two fractions into one fraction.
     
  10. Jan 14, 2013 #9
    Don't you still have to consider that x + 3 and x could be negative?
     
  11. Jan 14, 2013 #10

    HallsofIvy

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    Yes, it is really the same thing.

    Well, x+3< 0 is the same as x< -3 and, then, of course, x< 0.
    The three cases you want to consider are
    x< -3 when x+3< 0 and x< 0 are BOTH true.

    [itex]-3\le x< 0[/itex] when [itex]x+ 3\le 0[/itex] and x< 0.

    [itex]0\le x[/itex] when both x+3> 0 and [itex]x\ge 0[/itex].
     
  12. Jan 14, 2013 #11
    Yes, I quickly found out that when both are negative and when x+3<0 and x>0 there are no solutions. Maybe in the future I will have the instinct to realize that from the start. But for now I am just trying to get the medthod down.

    I got the same answer by putting both terms together and finding the commoj denomintor. Surely the answer must be correct.

    I thank you all for your help in rehabilitating my mathematics.
     
  13. Jan 15, 2013 #12
    You're not done yet.

    It's possible to establish the answer for the ranges {x<-3} and {x>0} just by considering function values relative to 1 (= x/x).

    However, the intermediate range {-3<x<0} is more complicated. Look at both function values at -2.9 and at -0.1 which should tell you there is more to discover.
     
    Last edited: Jan 15, 2013
  14. Jan 15, 2013 #13
    Are you supposed to state that there are asymptotes at -3 and 0 ?

    x cannot be -1/2 is that what you're referring to?
     
  15. Jan 15, 2013 #14

    SammyS

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    The resulting inequality is: [itex]\displaystyle \ \ \frac{-6x-3}{x(x+3)}<0\ .[/itex]

    This simplifies to: [itex]\displaystyle \ \ \frac{2x+1}{x(x+3)}>0\ .[/itex]

    You have three factors, one in the numerator and two in the denominator. Either all three must be positive, or one must be positive and two of them negative.
     
  16. Jan 15, 2013 #15
    There is another range for x which satisfies the condition.

    Both original expressions - and the combined expression too - are valid for ##x = -\frac 12##

    SammyS's simplified expression can also be expressed as $$ \frac{x+\frac 12}{x(x+3)}>0 $$
     
  17. Jan 15, 2013 #16
    It seems like I kept making careless errors. Thank you all for being patient with me.

    -3 < x < -1/2 , x > 0

    That should be the right answer.
     
  18. Jan 16, 2013 #17

    HallsofIvy

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    You want to determine when [itex](x-2)/(x+3)< (x+1)/(x)[/itex].

    As I said before, since we clearly want to multiply both sides by x+ 3 and x, to clear the fractions, we need to consider when they are positive of negative. So consider 3 cases:
    1) x< -3. Then both x+ 3 and x are negative. Multiplying the above inequality by x(x+ 3) is multiplying by a positive number (the product of two negatives) so the direction of inequality does not change. [itex](x- 2)(x)< (x+ 3)(x+ 1)[/itex]. [itex]x^2- 2x< x^2+ 4x+ 3[/itex]. Subtract [itex]x^2[/itex] from both sides to get [itex]-2x< 4x+ 3[/itex] so that [itex]0< 6x+ 3= 3(x+ 1)[/itex]. Dividing both sides by the positive number 3, we have [itex]0< x+ 1[/itex] or [itex]x< -1[/itex] which can't happen when x< -3.

    2) -3< x< 0. Now x+ 3 is positive but x is still negative. Multiplying both sides by x(x+ 3) is now multiplying by a negative number and changes the direction of the inequality: [itex](x- 2)(x)> (x+ 3)(x+ 1)[/itex]. The same calculations as before go through with the changed inequality sign: [itex]x< -2[/itex]. That tells us that the orignal inequality is true for [itex]-3< x< -2[/itex].

    3) x> 0. Now both x+ 3 and x are positive so j=multiplying both sides of the inequality by x(x+ 3) does not change the sign: [itex](x- 2)(x)< (x+ 3)(x+ 1)[/itex] and again we get [itex]x> -1[/itex]. Of course that is true for all x> 0 so we have the inequality true for all x> 0.

    We have, so far, that the inequality is true for -3< x< -2 and x> 0. You should also check to see if it is true at x= -3, x= -2, and x= 0.
     
    Last edited: Jan 17, 2013
  19. Jan 16, 2013 #18
    I think you made a mistake in the firsf case where you ended up with 3(x+2)

    Also the question didn't include any inequality symbols with equal signs. (No line under the inequality is what I mean.
     
  20. Jan 17, 2013 #19

    HallsofIvy

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    Oh, bother! Thanks. I have gone back and edited my post.
     
  21. Jan 17, 2013 #20
    And...
    ## (x−2)(x) > (x+3)(x+1) ##

    ## x^2-2x > x^2 +4x +3 ##

    ## -2x > 4x +3 ##

    ## 0 > 6x +3 ##

    ## x < -\frac 36 = -\frac 12 ##
     
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