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Simplifying expressions

  1. Sep 28, 2006 #1
    {(m^2 +2m -3)/(m^2-4)} / {(m^2 - 2m +1)/(m^2 -1)

    = {(m^2 +2m -3)/(m^2-4)} x {((m(m-1))/ (m^2 -2m +1)}

    = [(m+3)m] / [(m +2)(m-2)]

    does that like rightish? any opinions will be appreciated!

    ~Amy
     
  2. jcsd
  3. Sep 28, 2006 #2
    It's about 100% rightish.
     
  4. Sep 28, 2006 #3

    HallsofIvy

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    Not quite. m2- 1= (m-1)(m+1), not m(m-1)
    That's the only one you show factored, and it's wrong. It would be better to show how the other terms factor also.
     
  5. Sep 28, 2006 #4
    Sorry, missed that somehow.
     
  6. Sep 28, 2006 #5
    k thanks for the tip. my new answer is (m+2)/(m-1).. let me know if you need all the work shown (too lazy to type it if it's not really needed).

    here's another one:

    [3/(x+2)]+ [1/(x-1)]

    =[4x-1/(x+2)(x-1)]?


    and this one:
    [a-3/(a+3)(a-5)] - [a/(a +3)(a+4)]

    = (6a - 12)/ [(a+3)(a+4)(a-5)] ?

    also, i recently purchased a ti-83+ graphing calculator and some software called 'microsoft student 2007'. can any of these two help me check my calculations for this type of problem? i tried to use the microsoft student, but couldnt figure out who to simlify these times of problems and i havent the foggiest clue how to use my new graphing calculator. :eek:

    ~Amy
     
  7. Oct 5, 2006 #6
    Both look right to me.

    I've never owned or used a graphing calculator, so I'm afraid I can't help you with that. But there is a decent program out there called Autograph, that's quite a powerful yet simple-to-use tool. You can download a free 30 day trial of it from their website.

    I'm not totally sure, but I presume that if you enter in what's on the left hand side of your equation, and get it to draw the graph of that function, and then also get it to draw whats on the right hand side of the equation, you'd expect their graphs to overlap...if that makes sense lol.
     
  8. Oct 11, 2006 #7
    i've somewhat figured out my graphing calculator, like how to graph y= equations, but that's about it. there's a book on amazon about my specific caculator, so im going to buy that and see if it helps.

    ~Amy
     
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