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Math Help - solving equations

  1. #1
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    solving equations

    Please assist me in solving the following equation: (3t - 5) / (t - 1) = 2 + (2t) / (1 - t) Would t= -1 or 1 ? Thanks!
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  2. #2
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    Quote Originally Posted by jay1 View Post
    Please assist me in solving the following equation: (3t - 5) / (t - 1) = 2 + (2t) / (1 - t) Would t= -1 or 1 ? Thanks!
    You should be able to see straight away that t \neq 1 because it would create a 0 denominator.


    \frac{3t - 5}{t - 1} = 2 + \frac{2t}{1 - t}

    \frac{3t - 5}{t - 1} = 2 - \frac{2t}{t - 1}

    \frac{3t - 5}{t - 1} + \frac{2t}{t - 1} = 2

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

    5t - 5 = 2(t - 1)

    5t - 5 = 2t - 2

    3t = 3

    t = 1.



    So what this means is that this function will be almost identical to a different function - the only difference being that there is a point of discontinuity at t = 1. Because of this point of discontinuity, this equation does not have a solution.
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  3. #3
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    Quote Originally Posted by Prove It View Post
    You should be able to see straight away that t \neq 1 because it would create a 0 denominator.


    \frac{3t - 5}{t - 1} = 2 + \frac{2t}{1 - t}

    \frac{3t - 5}{t - 1} = 2 - \frac{2t}{t - 1}

    \frac{3t - 5}{t - 1} + \frac{2t}{t - 1} = 2

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

    5t - 5 = 2(t - 1)


    Okay, that's what was confusing me.. Thanks!

    5t - 5 = 2t - 2

    3t = 3

    t = 1.



    So what this means is that this function will be almost identical to a different function - the only difference being that there is a point of discontinuity at t = 1. Because of this point of discontinuity, this equation does not have a solution.

    Okay, that's what was confusing me.. Thanks!
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