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Math Help - Weird algebra

  1. #1
    Senior Member Paze's Avatar
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    Weird algebra

    (x+1)/(x-2)=3/(x-2)+5 - Wolfram|Alpha

    Why on earth can't I just multiply by (t-2) and get the solution t=2 ?!

    To make it absolutely clear that I should be able to just multiply by (t-2), I can rewrite this equation as:

    \frac{t+1}{t-2}=\frac{5t-7}{t-2}

    I am VERY confused..

    I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?
    Last edited by Paze; August 12th 2013 at 12:47 PM.
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  2. #2
    Senior Member Paze's Avatar
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    Re: Weird algebra

    I'm finding out that this is due to a problem in algebra called "extraneous solutions".

    Why does this problem arise? I'm having problems finding it... Is algebra flawed?
    Thanks from Shakarri
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  3. #3
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    Re: Weird algebra

    The reason that extraneous solutions sometimes occur is because if you were to raise both sides of the equation to a higher power such as cubed, you would increase the number of the solution. This new number night not satisfy the original equation.
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    Re: Weird algebra

    Quote Originally Posted by Paze View Post
    I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?
    Well it wouldn't give you the solution t=2 if you followed the rules of algebra and did not take \frac{0}{0}=1 (which is what you are doing when you cancel \frac{t-2}{t-2} to 1.
    Simply put this equation has no solution because it is wrong to begin with, you can see this more clearly if you rearrange it.

    \frac{t+1}{t-2}=\frac{5t-7}{t-2}

    \frac{t+1}{t-2}-\frac{5t-7}{t-2}=0

    \frac{1}{t-2}\times((t+1)-(5t-7))=0

    \frac{1}{t-2}\times(-4t+8)=0

    \frac{-4t+8}{t-2}=0

    \frac{-4(t-2)}{t-2}=0

    \frac{(t-2)}{t-2}=0

    Nothing divided by itself can equal 0 so you can see that the equation is false to begin with.
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  5. #5
    Senior Member Paze's Avatar
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    Re: Weird algebra

    Quote Originally Posted by Shakarri View Post
    Well it wouldn't give you the solution t=2 if you followed the rules of algebra and did not take \frac{0}{0}=1 (which is what you are doing when you cancel \frac{t-2}{t-2} to 1.
    Simply put this equation has no solution because it is wrong to begin with, you can see this more clearly if you rearrange it.

    \frac{t+1}{t-2}=\frac{5t-7}{t-2}

    \frac{t+1}{t-2}-\frac{5t-7}{t-2}=0

    \frac{1}{t-2}\times((t+1)-(5t-7))=0

    \frac{1}{t-2}\times(-4t+8)=0

    \frac{-4t+8}{t-2}=0

    \frac{-4(t-2)}{t-2}=0

    \frac{(t-2)}{t-2}=0

    Nothing divided by itself can equal 0 so you can see that the equation is false to begin with.
    I realize what the problem is now. However, I did not break any rules of algebra as you claim. Thanks though.
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    Re: Weird algebra

    Quote Originally Posted by Paze View Post
    (x+1)/(x-2)=3/(x-2)+5 - Wolfram|Alpha

    Why on earth can't I just multiply by (t-2) and get the solution t=2 ?!

    To make it absolutely clear that I should be able to just multiply by (t-2), I can rewrite this equation as:

    \frac{t+1}{t-2}=\frac{5t-7}{t-2}

    I am VERY confused..

    I do realize that t can't be 2 because of the denominator becoming 0, but why does it give me the algebraic solution t=2 and then it doesn't work?
    You were going to break a rule of algebra when you planned to multiply by t-2. If you look at it in steps you'll see that

    \frac{t+1}{t-2}=\frac{5t-7}{t-2}

    Multiply both sides by t-2

    \frac{t+1}{t-2}\times(t-2)=\frac{5t-7}{t-2}\times(t-2)

    \frac{t-2}{t-2}\times(t+1)=\frac{t-2}{t-2}\times(5t-7)

    (t+1)=(5t-7)
    At this point you have made the mistake of turning (t-2)/(t-2) into 1
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  7. #7
    Senior Member Paze's Avatar
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    Re: Weird algebra

    Quote Originally Posted by Shakarri View Post
    At this point you have made the mistake of turning (t-2)/(t-2) into 1
    I fail to see how dividing something by itself should not equal 1?
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    Re: Weird algebra

    It is not true for all t. When t=2 it is not equal to 1. So when you are trying to find a solution to the equation when you make the step to turn (t-2)/(t-2) into 1 you must make a note that any conclusions you reach are under the assumption that t is not equal to 2. And when you continued to find a solution you reached the conclusion that t=2. As I said "any conclusions you reach are under the assumption that t is not equal to 2." so your conclusion that t=2 is under the assumption that t is not equal to 2.
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  9. #9
    Senior Member Paze's Avatar
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    Re: Weird algebra

    Quote Originally Posted by Shakarri View Post
    It is not true for all t. When t=2 it is not equal to 1. So when you are trying to find a solution to the equation when you make the step to turn (t-2)/(t-2) into 1 you must make a note that any conclusions you reach are under the assumption that t is not equal to 2. And when you continued to find a solution you reached the conclusion that t=2. As I said "any conclusions you reach are under the assumption that t is not equal to 2." so your conclusion that t=2 is under the assumption that t is not equal to 2.
    That is correct, but I don't feel like I am breaking any "rules of algebra". I feel that the rules of algebra are leading me to a false conclusion due to the ALGEBRA expecting a higher degree equation...Is that crazy?
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