dividing fractions

**princess_21** · January 16th 2009, 08:39 PM

i got lost in complex fractions. i don't know how to input equations here..

**princess_21** · January 16th 2009, 09:59 PM

{1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?

**ronaldo_07** · January 16th 2009, 10:03 PM

You could let $1=\frac{m-1}{m-1}$ see if that helps.

**Grandad** · January 16th 2009, 10:05 PM

Hello princess_21

Originally Posted by princess_21

i got lost in complex fractions. i don't know how to input equations here..

Sorry, I can't understand your attached document. But have another look at the one you've already visited

http://www.mathhelpforum.com/math-he...fractions.html

If you're still really stuck, can you write one out neatly by hand, and either scan it or take a digital photo, and attach that instead?

Grandad

**Jhevon** · January 16th 2009, 10:15 PM

Originally Posted by princess_21

{1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?

is this it?

$\displaystyle \frac {1 - \frac 1{m - 1}}{1 + \frac 1{m - 1}} + \frac {1 - \frac 1{1 + \frac 1{m - 1}}}{1 - \frac 1{1 - \frac 1{m-1}}}$

**ronaldo_07** · January 16th 2009, 10:22 PM

Originally Posted by princess_21

{1 - [1/(m-1)] }/ {1+ [1/(m-1)]} + {1- [1/(1+(1/(m-1)]}/ {1-[1/(1-1/(m-1)}

is it stilll unclear?

$\frac{1-\frac{1}{m-1}}{1+\frac{1}{m-1}}+\frac{1-\frac{1}{1+\frac{1}{m-1}}}{1-\frac{1}{1-\frac{1}{m-1}}}$

Is this the correct equation?

**Jhevon** · January 16th 2009, 10:28 PM

Originally Posted by ronaldo_07

$\frac{1-\frac{1}{m-1}}{1+\frac{1}{m-1}}+\frac{1-\frac{1}{1+\frac{1}{m-1}}}{1-\frac{1}{1-\frac{1}{m-1}}}$

Is this the correct equation?

haha, well, both you and i interpreted it the same way. chances are we're correct...unless there was a typo.

would you like to proceed?

**ronaldo_07** · January 16th 2009, 10:35 PM

Originally Posted by ronaldo_07

You could let $1=\frac{m-1}{m-1}$ see if that helps.

I would let $1-\frac{1}{m-1}=\frac{m-1}{m-1}-\frac{1}{m-1}$

This simplyfies to (m-2)

Do this also for $1+\frac{1}{m-1}$ and it should come in a nice form for you to easily work out m

**princess_21** · January 16th 2009, 10:38 PM

here's another attachment..

**princess_21** · January 16th 2009, 10:40 PM

yeah. that's the question. thanks

**ronaldo_07** · January 16th 2009, 10:42 PM

Originally Posted by princess_21

here's another attachment..

We have interpreted correctly look above for help I have provided

**princess_21** · January 16th 2009, 10:45 PM

Originally Posted by Jhevon

haha, well, both you and i interpreted it the same way. chances are we're correct...unless there was a typo.

would you like to proceed?

yeah that's the question..

**princess_21** · January 16th 2009, 10:52 PM

i solved it this way.
let m= m-1

and then i got (2m-2)/(m+1)... is this correct?

**Jhevon** · January 16th 2009, 10:52 PM

Originally Posted by ronaldo_07

I would let $1-\frac{1}{m-1}=\frac{m-1}{m-1}-\frac{1}{m-1}$

This simplyfies to (m-2)

Do this also for $1+\frac{1}{m-1}$ and it should come in a nice form for you to easily work out m

indeed. just to flesh it out a bit

$\displaystyle \frac {1 - \frac 1{m - 1}}{1 + \frac 1{m - 1}} \cdot {\color{red} \frac {m - 1}{m - 1}} + \frac {1 - \frac 1{1 + \frac 1{m - 1}} \cdot {\color{red} \frac {m - 1}{m - 1}}}{1 - \frac 1{1 - \frac 1{m-1}} \cdot {\color{red} \frac {m - 1}{m - 1}}}$

$= \frac {m - 1 - 1}{m - 1 + 1} + \frac {1 - \frac {m - 1}{m - 1 + 1}}{1 - \frac {m - 1}{m - 1 - 1}}$

$= \frac {m - 2}m + \frac {1 - \frac {m - 1}m}{1 - \frac {m - 1}{m - 2}}$

now perform a similar trick on the second fraction as i did before and continue

**Jhevon** · January 16th 2009, 10:54 PM

Originally Posted by princess_21

i solved it this way.
let m= m-1

and then i got (2m-2)/(m+1)... is this correct?

you cannot solve it that way. you would be changing the value of the fraction. you would have to use another variable and then make the back-substitution afterwards. just follow the trick you have been shown. it is the easiest way

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