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Math Help - Solving rational equations

  1. #1
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    Solving rational equations

    I'm losing the will... I simply can't get my head around these rational equations:

    I've got the answers using MathCad, but I want to understand how they work... If someone wouldn't mind walking me through the process I would be very grateful. Please treat my like a child, as I can't for the life of me seem to grasp this concept. Also pretend that I don't understand LCM or GCF (I do for normal equations, but I don't seem able to transfer this knowledge to algebra).

    Here are a few examples I know the answers to, just not the workings:

    1) x+4/x - 14/2x-1 = 0. Answers are x=-1/2, x=4

    2) x+1/x - 4/2x-3 = 0. Answers are x=-1/2, x=3

    3) x/1-2x - 2/3x+1 = 0. Ansers are x=1/3, x=-2

    I would be insanely grateful if anyone could point me to some great algebra resources on the net, which have highly detailed step x step tutorials.

    Thanks a lot for your time,
    Mez
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  2. #2
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    Quote Originally Posted by mezhopking View Post
    I'm losing the will... I simply can't get my head around these rational equations:

    I've got the answers using MathCad, but I want to understand how they work... If someone wouldn't mind walking me through the process I would be very grateful. Please treat my like a child, as I can't for the life of me seem to grasp this concept. Also pretend that I don't understand LCM or GCF (I do for normal equations, but I don't seem able to transfer this knowledge to algebra).

    Here are a few examples I know the answers to, just not the workings:

    1) x+4/x - 14/2x-1 = 0. Answers are x=-1/2, x=4

    2) x+1/x - 4/2x-3 = 0. Answers are x=-1/2, x=3

    3) x/1-2x - 2/3x+1 = 0. Ansers are x=1/3, x=-2

    I would be insanely grateful if anyone could point me to some great algebra resources on the net, step x step in depth, highly detailed and explained tutorials.

    Thanks a lot for your time,
    Mez
    Getting a common denominator is the key in each case. 1):


    \frac{(x+4)(2x-1)}{x(2x-1)} - \frac{14x}{x(2x-1)} = 0


    \Rightarrow \frac{(x+4)(2x-1)-14x}{x(2x-1)} = 0


    \Rightarrow (x+4)(2x-1)-14x = 0

    \Rightarrow 2x^2 - 7x - 4 = 0

    \Rightarrow (x - 4)(2x + 1) = 0

    \Rightarrow x = 4, ~ -\frac{1}{2}.
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