Results 1 to 3 of 3

Thread: solving equations

  1. #1
    Member
    Joined
    Dec 2009
    Posts
    122

    solving equations

    Please assist me in solving the following equation: (3t - 5) / (t - 1) = 2 + (2t) / (1 - t) Would t= -1 or 1 ? Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946
    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 $\displaystyle t \neq 1$ because it would create a $\displaystyle 0$ denominator.


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

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

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

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

    $\displaystyle 5t - 5 = 2(t - 1)$

    $\displaystyle 5t - 5 = 2t - 2$

    $\displaystyle 3t = 3$

    $\displaystyle t = 1$.



    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 $\displaystyle t = 1$. Because of this point of discontinuity, this equation does not have a solution.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Dec 2009
    Posts
    122
    Quote Originally Posted by Prove It View Post
    You should be able to see straight away that $\displaystyle t \neq 1$ because it would create a $\displaystyle 0$ denominator.


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

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

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

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

    $\displaystyle 5t - 5 = 2(t - 1)$


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

    $\displaystyle 5t - 5 = 2t - 2$

    $\displaystyle 3t = 3$

    $\displaystyle t = 1$.



    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 $\displaystyle t = 1$. Because of this point of discontinuity, this equation does not have a solution.

    Okay, that's what was confusing me.. Thanks!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Nov 30th 2011, 01:41 AM
  2. Solving two equations
    Posted in the Algebra Forum
    Replies: 7
    Last Post: Nov 21st 2011, 02:55 PM
  3. Solving Equations
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Sep 26th 2009, 12:29 PM
  4. Solving Equations..please Help!
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Mar 18th 2008, 08:47 AM
  5. solving equations
    Posted in the Algebra Forum
    Replies: 3
    Last Post: May 16th 2006, 03:58 PM

Search Tags


/mathhelpforum @mathhelpforum