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

1. ## 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.

2. Originally Posted by paulbk108
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?

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

next time, post the equation in the body of the post rather than the title only.

3. $\displaystyle \displaystyle \frac{a}{A}+\frac{b}{B}+\frac{c}{C}+\frac{d}{D}+ . . . . . = \frac {a\cdot(B\cdot C\cdot D \cdot ....) +b\cdot (A\cdot C\cdot D \cdot ....) + c\cdot (B\cdot A\cdot D \cdot ....) +d\cdot (B\cdot C\cdot A \cdot ....) + .... }{A\cdot B\cdot C\cdot D \cdot ....}$

4. 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.

5. Originally Posted by paulbk108
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.

6. Is $\displaystyle s^2$ or $\displaystyle s^2+1$ the denominator of the right-hand side ?

7. 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.

8. Originally Posted by paulbk108
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.
$\displaystyle \displaystyle 1+\frac {s+2}{s} +\frac {4}{s(s-1)} =$

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

multiply all with " s(s-1) "

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

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

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

or i misunderstand the question

9. Originally Posted by paulbk108
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 ...

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

if you really mean this ...

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

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

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

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

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

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

you're done.

10. 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?

11. Originally Posted by paulbk108
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" ...

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

12. Of course! Stupid me!

Thanks so much...