# Looking for Equation Solver

Discussion in 'Perl Misc' started by Chris, Sep 3, 2005.

1. ### ChrisGuest

Hi,

does anybody know a Perl program or script that is able to solve
mathematical equations. So, if I input a string, say, \$e = "2x+1=5";
the output would be "x=2".

If anybody knows about a program that can do something like this, please
let me know.

Thank you,
Chris

Chris, Sep 3, 2005

2. ### Ignoramus14363Guest

I wrote one

http://www.algebra.com/services/rendering/simplifier.mpl

Besides simplifying, it shows work and plots math formulas. It is not
available publicly though. There is a Math::Symbolic module, but I
found it extremely lacking.

i

On Sat, 03 Sep 2005 19:49:43 +0200, Chris <no@spam> wrote:
> Hi,
>
> does anybody know a Perl program or script that is able to solve
> mathematical equations. So, if I input a string, say, \$e = "2x+1=5";
> the output would be "x=2".
>
> If anybody knows about a program that can do something like this, please
> let me know.
>
> Thank you,
> Chris

--

Ignoramus14363, Sep 3, 2005

3. ### Guest

Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> I wrote one
>
> http://www.algebra.com/services/rendering/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.

Xho

--
Usenet Newsgroup Service \$9.95/Month 30GB

, Sep 3, 2005
4. ### Ignoramus14363Guest

On 03 Sep 2005 22:51:13 GMT, <> wrote:
> Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
>> I wrote one
>>
>> http://www.algebra.com/services/rendering/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

Ignoramus14363, Sep 4, 2005
5. ### Keith KellerGuest

On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
>
> On 03 Sep 2005 22:51:13 GMT, <> 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! :-\

--keith

--
-francisco.ca.us
(try just my userid to email me)
AOLSFAQ=http://wombat.san-francisco.ca.us/cgi-bin/fom
see X- headers for PGP signature information

Keith Keller, Sep 4, 2005
6. ### Ignoramus14363Guest

On Sat, 3 Sep 2005 22:47:50 -0700, Keith Keller <-francisco.ca.us> wrote:
> On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
>>
>> On 03 Sep 2005 22:51:13 GMT, <> 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

Ignoramus14363, Sep 4, 2005
7. ### BZGuest

Ignoramus14363 wrote in comp.lang.perl.misc:
> > "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?

(x-3)^2/(x-4)^2

--
BZ

BZ, Sep 4, 2005
8. ### Paul LalliGuest

OT: Math Simplification (WAS: Looking for Equation Solver)

Ignoramus14363 wrote:
> Keith Keller wrote:
> > Ignoramus14363 wrote:
> >> 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.
> >>

> If you can explain why the above simplification is incorrect, I would
> appreciate hearing that.
>

Uhm. Because it's not mathematically valid? Try any value of X. The
original does not produce the same result as the simplification. (Hint
- the easiest value to plug in would be 0, which gives -3/2 for the
simplification, but
(-9/10) / (6/15) = (-9/10) * (15/6) = (-3/2) * (3/2) = -9/4 for the
original).

fwiw, when I do it on paper I get:
(x^2 - 6x - 9) / (x^2 -4)

(Granted, it has been about 5 years since I last took a math class...)

Paul Lalli

Paul Lalli, Sep 4, 2005
9. ### DaveGuest

"BZ" <> wrote in message
news:...
> Ignoramus14363 wrote in comp.lang.perl.misc:
>> > "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?

>
> (x-3)^2/(x-4)^2
>
> --
> BZ

No, Paul is right. It can also be written:
((x-3)^2)/((x+2)(x-2))
or
(x-3)^2/(x^2-4)
the latter of which is perhaps what BZ meant to type.

Dave, Sep 4, 2005
10. ### DaveGuest

OT: Re: Looking for Equation Solver

"Ignoramus14363" <ignoramus14363@NOSPAM.14363.invalid> wrote in message
news:bdwSe.98947\$...
> On Sat, 3 Sep 2005 22:47:50 -0700, Keith Keller
> <-francisco.ca.us> wrote:
>> On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid>
>> wrote:
>>>
>>> On 03 Sep 2005 22:51:13 GMT, <>
>>> 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

Dave, Sep 4, 2005
11. ### Guest

Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> On 03 Sep 2005 22:51:13 GMT, <> wrote:
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> >> I wrote one
> >>
> >> http://www.algebra.com/services/rendering/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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Usenet Newsgroup Service \$9.95/Month 30GB

, Sep 5, 2005
12. ### Ignoramus25850Guest

Re: OT: Re: Looking for Equation Solver

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 <> wrote:
>
> "Ignoramus14363" <ignoramus14363@NOSPAM.14363.invalid> wrote in message
> news:bdwSe.98947\$...
>> On Sat, 3 Sep 2005 22:47:50 -0700, Keith Keller
>> <-francisco.ca.us> wrote:
>>> On 2005-09-03, Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid>
>>> wrote:
>>>>
>>>> On 03 Sep 2005 22:51:13 GMT, <>
>>>> 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
>
>

--

Ignoramus25850, Sep 7, 2005
13. ### Peyton BlandGuest

Hi,

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

In article <20050904211634.239\$>, <>
wrote:

> Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > On 03 Sep 2005 22:51:13 GMT, <> wrote:
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > >> I wrote one
> > >>
> > >> http://www.algebra.com/services/rendering/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

Peyton Bland, Nov 14, 2005
14. ### Paul LalliGuest

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\$>, <>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, <> 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

Paul Lalli, Nov 14, 2005
15. ### Guest

Peyton Bland <> 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\$>, <>
> wrote:
>
> > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > On 03 Sep 2005 22:51:13 GMT, <>
> > > wrote:
> > > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > >> I wrote one
> > > >>
> > > >> http://www.algebra.com/services/rendering/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

--
Usenet Newsgroup Service \$9.95/Month 30GB

, Nov 14, 2005
16. ### Peyton BlandGuest

In article <>, Paul
Lalli <> 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\$>, <>
> > wrote:
> >
> > > Ignoramus14363 <ignoramus14363@NOSPAM.14363.invalid> wrote:
> > > > On 03 Sep 2005 22:51:13 GMT, <>
> > > > 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

Peyton Bland, Nov 15, 2005