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Thread: Rational inequality word problem

  1. #1
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    Question Rational inequality word problem

    Hi,

    I hope someone can help. I'm trying to understand in what cases it is okay for me to remove the denominator from a rational inequality without having to flip the inequality sign. I thought it was okay in the event that I knew that the denominator was positive. However the following word problem appears to prove otherwise...

    A(t) = \frac{360}{t+6}
    C(t) = \frac{50}{41-2t}


    I'm trying to determine when A(t) > C(t). Both functions measure population size over "t" years starting at t=0. Since I know that the denominator is positive, this means that I can remove the denominator without having to flip the inequality sign.

    I got the solution which was A(t) > C(t) when t < 18.78. However, I know that this is only one of my solutions because when I compare the graphs using graphing technology, I also notice that A(t) > C(t) when t > 20.5. Can someone please explain to me why this solution did not show up as part of my solution set?
    Last edited by otownsend; Jul 7th 2017 at 12:03 PM.
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  2. #2
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    Re: Rational inequality word problem

    Never mind I think I just figured out my own question. Even though t > 0 the denominator could very well be of a negative value for C(t).
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  3. #3
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    Re: Rational inequality word problem

    You want to solve \frac{360}{t+ 6}> \frac{50}{41- 2t}. The first thing you would like to do is get rid of the fractions by multiplying both sides by the denominators, t+ 6 and 41- 2t, but multiplying by a negative number changes the direction of the inequality. We first need to determine where each is negative and where positive. t+ 6< 0 if and only if t< -6. 41- 2t< 0 if and only if 41< 2t so 41/2< t. Those two points divide the number line into 3 intervals.

    If t< -6 then t+ 6 is negative but 41- 2t is positive. Multiplying by the two denominators we have multiplied by a negative number once so must change to direction of the inequality: 360(41- 2t)< 50(t+ 6).

    If -6< t< 41/2 then t+ 6 and 41- 2t are both positive: 360(41- 2t)> 50(t+ 6).

    If t> 41/2 then t+ 6 is positive but 41- 2t is negative: 360(41- 2t)< 50(t+ 6).
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  4. #4
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    Re: Rational inequality word problem

    Solve each of those and, of course, take the intersection of the solution set with the interval on which that inequality is the valid one.
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