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Math Help - 1 + [(s+2)/s] + [4/s(s-1)] = 4s(s+20/s^2 + 1

  1. #1
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    1 + [(s+2)/s] + [4/s(s-1)] = 4s(s+20/s^2 + 1

    In a previous thread I was shown the solution to a similar problem - using the rule of a/m + b/n = (an + bm)/mn. To my knowledge this is only applicable when there are two fractions.

    I would like to see the simplification process of this.

    Can anyone help?

    Regards.
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  2. #2
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    Quote Originally Posted by paulbk108 View Post
    In a previous thread I was shown the solution to a similar problem - using the rule of a/m + b/n = (an + bm)/mn. To my knowledge this is only applicable when there are two fractions.

    I would like to see the simplification process of this.

    Can anyone help?

    Regards.
    is this your equation?

    1 + \frac{s+2}{s} + \frac{4}{s(s-1)} = \frac{4s(s+20)}{s^2+1}

    next time, post the equation in the body of the post rather than the title only.
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  3. #3
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     \displaystyle \frac{a}{A}+\frac{b}{B}+\frac{c}{C}+\frac{d}{D}+ . . . . . = \frac {a\cdot(B\cdot C\cdot D \cdot ....) +b\cdot (A\cdot C\cdot D \cdot ....) + c\cdot (B\cdot A\cdot D \cdot ....) +d\cdot (B\cdot C\cdot A \cdot ....) + .... }{A\cdot B\cdot C\cdot D \cdot ....}
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  4. #4
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    My apologies - thought it would grab more attention in the title.

    The equation is 1 + s+2/s + 4/s(s-1) = 4s(s+2)/S^2+1. So yes it is the equation - I just mucked up with a bracket towards the end.

    Regards.
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  5. #5
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    Quote Originally Posted by paulbk108 View Post
    My apologies - thought it would grab more attention in the title.

    The equation is 1 + s+2/s + 4/s(s-1) = 4s(s+2)/S^2+1. So yes it is the equation - I just mucked up with a bracket towards the end.

    Regards.
    Don't know where you dug up this equation, but fyi, it works out to a rather nasty quartic equation that requires technology to solve (two irrational real roots). If you are trying to learn the nuances of algebra. I recommend you pick an equation that's a bit easier to work.
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  6. #6
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    Is s^2 or s^2+1 the denominator of the right-hand side ?
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  7. #7
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    My apologies folks - I misread the equation. I have a denominator in an equation that is 1 + s+2/s + 4/s(s-1). I would like to see this simplified. I have done this stuff to the death in a past life - it has been a while that's all and I'm still cleaning off the rust :-)

    Sorry to trouble you all, and thanks so far.

    Regards.
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  8. #8
    Senior Member yeKciM's Avatar
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    Quote Originally Posted by paulbk108 View Post
    My apologies folks - I misread the equation. I have a denominator in an equation that is 1 + s+2/s + 4/s(s-1). I would like to see this simplified. I have done this stuff to the death in a past life - it has been a while that's all and I'm still cleaning off the rust :-)

    Sorry to trouble you all, and thanks so far.

    Regards.
     \displaystyle 1+\frac {s+2}{s} +\frac {4}{s(s-1)}  =

     \displaystyle = 1+ 1 +\frac {2}{s} + \frac {4}{s(s-1)} =

    multiply all with " s(s-1) "

     \displaystyle = 2s(s-1)+2(s-1) + 4  =

     \displaystyle = 2s^2-2s+2s-2+4 =

     \displaystyle = 2s^2+2 = 2(s^2+1)

    or i misunderstand the question
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  9. #9
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    Quote Originally Posted by paulbk108 View Post
    My apologies folks - I misread the equation. I have a denominator in an equation that is 1 + s+2/s + 4/s(s-1). I would like to see this simplified. I have done this stuff to the death in a past life - it has been a while that's all and I'm still cleaning off the rust :-)

    Sorry to trouble you all, and thanks so far.

    Regards.
    this is what you have posted ...

    1 + s + \frac{2}{s} + \frac{4}{s} (s-1)

    if you really mean this ...

    1 + \frac{s+2}{s} + \frac{4}{s(s-1)} , then you should type it out thus ...

    1 + (s+2)/s + 4/[s(s-1)]


    so ... if this is the expression, then the common denominator is s(s-1)

    \frac{s(s-1)}{s(s-1)} + \frac{(s+2)(s-1)}{s(s-1)} + \frac{4}{s(s-1)}

    \frac{s(s-1) + (s+2)(s-1) + 4}{s(s-1)}

    \frac{s^2 - s + s^2 + s - 2 + 4}{s(s-1)}

    \frac{2s^2 + 2}{s(s-1)}

    \frac{2(s^2+1)}{s(s-1)}

    you're done.
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  10. #10
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    Much appreciated.

    One small question: in the second term why was the (s+2) not multiplied by the s(s-1) and only the (s-1) if s(s-1) is the common denominator?
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  11. #11
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    Quote Originally Posted by paulbk108 View Post
    Much appreciated.

    One small question: in the second term why was the (s+2) not multiplied by the s(s-1) and only the (s-1) if s(s-1) is the common denominator?
    because the denominator already possessed an "s" ...

    \frac{s+2}{s} \cdot \frac{s-1}{s-1} = \frac{(s+2)(s-1)}{s(s-1)}
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  12. #12
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    Of course! Stupid me!

    Thanks so much...
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