Velocity Reviews > Perl > Looking for Equation Solver

# Looking for Equation Solver

xhoster@gmail.com
Guest
Posts: n/a

 09-05-2005
Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> On 03 Sep 2005 22:51:13 GMT, http://www.velocityreviews.com/forums/(E-Mail Removed) <(E-Mail Removed)> wrote:
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> >> I wrote one
> >>
> >> http://www.algebra.com/services/rend...simplifier.mpl
> >>
> >> Besides simplifying, it shows work and plots math formulas. It is not
> >> available publicly though.

> >
> > Also, it doesn't give the right answer.
> >
> > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > simplifies to (x-3)/(x+2)"
> >
> > No, it doesn't.

>
> I will appreciate corrections and suggestions. What's the right answer
> there?

I didn't solve it fully; I just set x to zero which was enough to show the
simplification was wrong.

It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
your program realizes that (x-3) is a common factor in two of the quadratic
groups, but then "cancels out" both of those groups in their entirety,
rather than just the factor of x-3.

Xho

>
> i

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Ignoramus25850
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Posts: n/a

 09-07-2005
Thanks guys (all of you). You got it. I corrected the solver, it was a
programming mistake. Thank you!

i

On Sun, 4 Sep 2005 21:29:30 +0200, Dave <(E-Mail Removed)> wrote:
>
> "Ignoramus14363" <ignoramus14363@NOSPAM.14363.invalid> wrote in message
> news:bdwSe.98947\$(E-Mail Removed).. .
>> On Sat, 3 Sep 2005 22:47:50 -0700, Keith Keller
>> <(E-Mail Removed)-francisco.ca.us> wrote:
>>> On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid>
>>> wrote:
>>>>
>>>> On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
>>>> wrote:
>>>>>
>>>>> Also, it doesn't give the right answer.
>>>>>
>>>>> "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
>>>>> simplifies to
>>>>> (x-3)/(x+2)"
>>>>>
>>>>> No, it doesn't.
>>>>
>>>> I will appreciate corrections and suggestions. What's the right answer
>>>> there?
>>>
>>> If you can't figure it out on paper, why are you writing a program to do
>>> it? Sheesh, I might as well go write an OS! :-\

>>
>> Let me know when you have something useful and substantial to say,
>> okay?
>>
>> (hint, I did solve it on paper)
>>
>> If you can explain why the above simplification is incorrect, I would
>> appreciate hearing that.
>>
>> i
>>

>
> The error in the explanation this example is at the line:
> 'Canceled out common factors (x-(3)),(x-(3))'
>
> Where (1) is left in each case instead of (x-3) and (x-2) in numerator and
> denominator respectively.
>
> The explantion would be clearer if you showed the factorisation before the
> cancellation, and this error would have been more obvious too.
>
> Hope this helps
>
> Dave
>
>

--

Peyton Bland
Guest
Posts: n/a

 11-14-2005

Hi,

I'm coming to the party a bit late, but you guys are scaring me, so I

In article <20050904211634.239\$(E-Mail Removed)>, <(E-Mail Removed)>
wrote:

> Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)> wrote:
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > >> I wrote one
> > >>
> > >> http://www.algebra.com/services/rend...simplifier.mpl
> > >>
> > >> Besides simplifying, it shows work and plots math formulas. It is not
> > >> available publicly though.
> > >
> > > Also, it doesn't give the right answer.
> > >
> > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > simplifies to (x-3)/(x+2)"
> > >
> > > No, it doesn't.

> >
> > I will appreciate corrections and suggestions. What's the right answer
> > there?

>
> I didn't solve it fully; I just set x to zero which was enough to show the
> simplification was wrong.

I did this, too, and got -1 when x = 0. What did you get, Xho?

> It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> your program realizes that (x-3) is a common factor in two of the quadratic
> groups, but then "cancels out" both of those groups in their entirety,
> rather than just the factor of x-3.

Have another look... I get (x+2)/(x-2) . Hopefully, no one coerced
the software in question to yield any of these other answers given
earlier.

Cheers,
Peyton Bland

Paul Lalli
Guest
Posts: n/a

 11-14-2005
Peyton Bland wrote:
> I'm coming to the party a bit late, but you guys are scaring me, so I

It may have been wise to figure out what you're talking about before
reviving a 3 month old thread.

> In article <20050904211634.239\$(E-Mail Removed)>, <(E-Mail Removed)>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)> wrote:
> > > >
> > > > Also, it doesn't give the right answer.
> > > >
> > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > simplifies to (x-3)/(x+2)"
> > > >
> > > > No, it doesn't.
> > >
> > > I will appreciate corrections and suggestions. What's the right answer
> > > there?

> >
> > I didn't solve it fully; I just set x to zero which was enough to show the
> > simplification was wrong.

>
> I did this, too, and got -1 when x = 0. What did you get, Xho?

My guess would be the correct answer that we all gave three months ago.
-9/4

((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))

(0^2-6*0+9)/(0^2-3*0-10)
------------------------
(0^2-5*0+6)/(0^2-8*0+15)

9 / -10
------
6 / 15

9 15
- -- * --
10 6

135 27 9
- --- = - -- = - -
60 12 4

(make sure you view the above in a fixed-width font)

> > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> > your program realizes that (x-3) is a common factor in two of the quadratic
> > groups, but then "cancels out" both of those groups in their entirety,
> > rather than just the factor of x-3.

>
> Have another look... I get (x+2)/(x-2) .

Yes, well, you're wrong. See previous responses for how and why.

> Hopefully, no one coerced
> the software in question to yield any of these other answers given
> earlier.

Yes, improving the buggy software would certainly be worse than using a
working one to begin with.

Paul Lalli

xhoster@gmail.com
Guest
Posts: n/a

 11-14-2005
Peyton Bland <(E-Mail Removed)> wrote:
> Hi,
>
> I'm coming to the party a bit late, but you guys are scaring me, so I

That being the case, you might want to spend some more time explaining
what it is you are trying to say. It is hard to just pick up a thread
after 3 months.

> In article <20050904211634.239\$(E-Mail Removed)>, <(E-Mail Removed)>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
> > > wrote:
> > > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > >> I wrote one
> > > >>
> > > >> http://www.algebra.com/services/rend...simplifier.mpl
> > > >>
> > > >> Besides simplifying, it shows work and plots math formulas. It is
> > > >> not available publicly though.
> > > >
> > > > Also, it doesn't give the right answer.
> > > >
> > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > simplifies to (x-3)/(x+2)"
> > > >
> > > > No, it doesn't.
> > >
> > > I will appreciate corrections and suggestions. What's the right

> >
> > I didn't solve it fully; I just set x to zero which was enough to show
> > the simplification was wrong.

>
> I did this, too, and got -1 when x = 0. What did you get, Xho?

Well, for the original full form, I got -9/4. For the original incorrect
simplification I got -3/2. For the correct simplification I got
(obviously) -9/4. For your new incorrect simplification I got -1.

>
> > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem
> > like your program realizes that (x-3) is a common factor in two of the
> > quadratic groups, but then "cancels out" both of those groups in their
> > entirety, rather than just the factor of x-3.

>
> Have another look...

Are you trying to tell me to double check my work, or are you telling me
that the bug has been fixed and I should go look at the original web page
again?

> I get (x+2)/(x-2) .

Which is obiously wrong, as (setting x=0) -9/4 != -1.

> Hopefully, no one coerced
> the software in question to yield any of these other answers given
> earlier.

I have no idea what that means.

Xho

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Peyton Bland
Guest
Posts: n/a

 11-15-2005
In article <(E-Mail Removed). com>, Paul
Lalli <(E-Mail Removed)> wrote:

> Peyton Bland wrote:
> > I'm coming to the party a bit late, but you guys are scaring me, so I

>
> It may have been wise to figure out what you're talking about before
> reviving a 3 month old thread.

My sincere apologies to the group. I carelessly mis-read the
expression (the ratio of polynomials).

> > In article <20050904211634.239\$(E-Mail Removed)>, <(E-Mail Removed)>
> > wrote:
> >
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > > On 03 Sep 2005 22:51:13 GMT, (E-Mail Removed) <(E-Mail Removed)>
> > > > wrote:
> > > > >
> > > > > Also, it doesn't give the right answer.
> > > > >
> > > > > "Text form: ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
> > > > > simplifies to (x-3)/(x+2)"
> > > > >
> > > > > No, it doesn't.
> > > >
> > > > I will appreciate corrections and suggestions. What's the right answer
> > > > there?
> > >
> > > I didn't solve it fully; I just set x to zero which was enough to show the
> > > simplification was wrong.

> >
> > I did this, too, and got -1 when x = 0. What did you get, Xho?

>
> My guess would be the correct answer that we all gave three months ago.
> -9/4
>
> ((x^2-6x+9)/(x^2-3x-10))/((x^2-5x+6)/(x^2-8x+15))
>
>
> (0^2-6*0+9)/(0^2-3*0-10)
> ------------------------
> (0^2-5*0+6)/(0^2-8*0+15)
>
> 9 / -10
> ------
> 6 / 15
>
> 9 15
> - -- * --
> 10 6
>
> 135 27 9
> - --- = - -- = - -
> 60 12 4
>
>
> (make sure you view the above in a fixed-width font)
>
> > > It appears that the correct answer is (x-3)^2/(x-2)/(x+2). It seem like
> > > your program realizes that (x-3) is a common factor in two of the
> > > groups, but then "cancels out" both of those groups in their entirety,
> > > rather than just the factor of x-3.

> >
> > Have another look... I get (x+2)/(x-2) .

>
> Yes, well, you're wrong. See previous responses for how and why.
>
> > Hopefully, no one coerced
> > the software in question to yield any of these other answers given
> > earlier.

>
> Yes, improving the buggy software would certainly be worse than using a
> working one to begin with.
>
> Paul Lalli

Peyton Bland