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Thread: Trying to go back to school to learn calculus (x-2) / (x+4) <7

  1. #1
    Nov 2010

    Question Trying to go back to school to learn calculus (x-2) / (x+4) <7

    Ok, I know this problem is an algebra one and not a calculus but the issue is that I am trying to get back into maths (mature age student here) and have bought an introductory calculus book. In reading the initial chapters it gave some examples of algebra that we should be able to easily handle to be able to proceed further with calculus study (quadratics etc).

    One example I could not handle was this
    (X-2) < 7

    It is not that I need the answer but need to find out how to process questions like this. I am sure it is a part of my algebra studies but can't remember, the words polynomial is jumping up from the back of my memory. Can anyone show me the working or tell me which area I should be looking at (perhaps an online tutorial on this.)

    Really I don't know where to start but will try some working (didn't want to post here without at least trying myself)
    (x-2) < 7

    = (x-2) < 7 (x+4)
    = (x-2) < 7x + 28
    = x < 7x + 28 + 2
    = x < 7x + 30
    = 0 < 6x +30
    = -30 < 6x
    = -5 < x

    Is this correct?
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  2. #2
    Junior Member
    Sep 2012
    Due North

    Re: Trying to go back to school to learn calculus (x-2) / (x+4) <7

    Try a few numbers. Can x be -4 ?
    What about -10 ?
    There are 2 points of x has a range of values that it can be greater than, and a range where x has to be less than...
    -5 is part of the answer but is it x > -5 ?
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  3. #3
    MHF Contributor
    Sep 2012
    Washington DC USA

    Re: Trying to go back to school to learn calculus (x-2) / (x+4) <7

    Your algebra is correct, but you overlooked something that resulted in a wrong answer. To see that it's incorrect, look at x=-4.5. Then x>-5, but (x-2)/(x+4) works out to be (-6.5)/(-.5) = 13, which is not less than 7. The thing you missed is that when you multiply both sides of an inequality, the inequality will "flip" if the number you're multiplying by is negative. This is a common headache when doing algebra with inequalities, and so is something to always keep an eye out for whenever dealing with inequalities. When you multiply inequalities, always pause and ask if what you're multiplying by is positive or not. This is similar to fractions, where whenever you have a fraction, you need to pause and ask yourself if you might be dividing by 0.

    There are many ways to solve this problem. Here's how I'd do it: $\displaystyle \frac{x-2}{x+4}<7$

    $\displaystyle \Leftrightarrow \frac{x-2}{x+4}-7<0 \Leftrightarrow \frac{x-2}{x+4}-\frac{7(x+4)}{x+4}<0 \Leftrightarrow \frac{(x-2) - (7x+28)}{x+4} < 0$

    $\displaystyle \Leftrightarrow \frac{-6x-30}{x+4} < 0 \Leftrightarrow (-6)(\frac{x+5}{x+4}) < 0 \Leftrightarrow \frac{x+5}{x+4} > 0 \Leftrightarrow (x+5)(\frac{1}{x+4}) > 0$.

    Note the flip when divided by -6. Now use that a product is positive means that either both factors are positive or both factors are negative.

    Thus EITHER $\displaystyle \{ x+5>0$ AND $\displaystyle \frac{1}{x+4} > 0 \}$ OR $\displaystyle \{ x+5<0$ AND $\displaystyle \frac{1}{x+4}<0 \}$.

    Now notice that, for any non-zero real number u, $\displaystyle \frac{1}{u} < 0 \Leftrightarrow u<0$, and likewise, $\displaystyle \frac{1}{u} > 0 \Leftrightarrow u>0$.

    Thus EITHER $\displaystyle \{ x+5>0$ AND $\displaystyle x+4>0 \}$ OR $\displaystyle \{ x+5<0$ AND $\displaystyle x+4<0 \}$.

    For the first part, $\displaystyle \{ x+5>0$ AND $\displaystyle x+4>0 \}$ means that $\displaystyle \{ x>-5$ AND $\displaystyle x>-4 \}$, which only happens when x>-4.

    For the second part, $\displaystyle \{ x+5<0$ AND $\displaystyle x+4<0 \}$ means that $\displaystyle \{ x<-5$ AND $\displaystyle x<-4 \}$, which only happens when x<-5.

    Thus "either x<-5 or x>-4" is the answer. This could also be described as x not in the interval [-5, -4], or as $\displaystyle (-\infty,-5) \cup (-4, \infty)$.


    Another way to do it is to follow the algebra you did, but split into cases where x+4 is positive and negative, so that you'll know whether or not to flip the inequality.
    Case x+4>0. Then ... (your work)... , so x>-5. But in the case x>-4, the "solution" x>-5 amounts to only that x>-4. Thus x>-4 is part of the solution.
    Case x+4<0. Then ... (your work, BUT with the inequality flipped when multiplied by (x+4))... , so x<-5. Thus in the case x<-4, the "solution" is x<-5, which means that x<-5 for this case.
    Thus the complete solution is any x such that x>-4 or x<-5.


    Another way is solve for where that fraction = 7 (getting x=-5). Then use that whenever a function is continuous on an interval, and you know where it's 7, then *between* where it's 7, it's either always greater than 7, or always less than 7. Thus you can check if it's more or less than 7 at a single point, and that will tell you the inequality over the whole interval. In this case, the function fails to be continuous only at x = -4, and is 7 only at x=-5. thus the line is split into intervals x<-5, -5<x<-4, and x>-4, and on each of those intervals, the function is always either greater than 7, or less than 7.
    For x>-4: Plug in x = 0, and you get (0-2)/(0+4) = -1/2 < 7. Thus the function is less than 7 for all x >-4.
    For -5<x<-4: Plug in x=-4.5, and you get 13 > 7. Thus the function is greater than 7 when -5<x<-4.
    For x<-5: Plug in x = -6, and you get (-6+2)/(-6+4) = (-4)/(-2) = 2 < 7. Thus the function is less than 7 for all x < -5.
    Last edited by johnsomeone; Sep 12th 2012 at 12:37 AM.
    Thanks from Blafin
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  4. #4
    Nov 2010

    Re: Trying to go back to school to learn calculus (x-2) / (x+4) <7

    Johnsomeone, thanks for the help, highly appreciated .......[clicks thanks]
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