Simultaneous Equations

pmb · Joined: 16 Aug 2007 Posts: 1

It's a long time since I was at school, and I've forgotten a lot. But I'm
trying to solve three equations. They look like they're solvable, but I
can't do it. It's possible that they contain some absurdity that makes them
unsolvable. But here they are:

a=10+b+c
2b=10+a+c
3c=10+a+b

Is there as solution?

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Archived from group: uk>education>maths

Martin · Joined: 05 Aug 2007 Posts: 22

"pmb" wrote in message @bt.com...
> It's a long time since I was at school, and I've forgotten a lot. But I'm
> trying to solve three equations. They look like they're solvable, but I
> can't do it. It's possible that they contain some absurdity that makes
> them unsolvable. But here they are:
>
> a=10+b+c
> 2b=10+a+c
> 3c=10+a+b
>
> Is there as solution?
>

Yes. What have you done so far, and where did you get stuck?

(Tip - take care to distinguish positive and negative values as you work
through it....)

--
Martin
[ remove barrier to reply ]

Stan Brown · Joined: 05 Aug 2007 Posts: 233

Sun, 12 Aug 2007 13:05:48 +0100 from pmb :
> It's a long time since I was at school, and I've forgotten a lot.

Don't post the same question multiple times. That just wastes
everyone's time. I've already posted a response in alt.algebra.help.

--
"The Internet is famously powered by the twin engines of
bitterness and contempt." -- Nathan Rabin, /The Onion/
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
http://OakRoadSystems.com/

terpsichore334692 · Joined: 16 Aug 2007 Posts: 3

On Aug 12, 2:14 pm, "Martin" wrote:
> "pmb" wrote in message
>
> @bt.com...
>
> > It's a long time since I was at school, and I've forgotten a lot. But I'm
> > trying to solve three equations. They look like they're solvable, but I
> > can't do it. It's possible that they contain some absurdity that makes
> > them unsolvable. But here they are:
>
> > a=10+b+c
> > 2b=10+a+c
> > 3c=10+a+b
>
> > Is there as solution?
>
> Yes. What have you done so far, and where did you get stuck?
>
> (Tip - take care to distinguish positive and negative values as you work
> through it....)

Well, it was getting a negative answer that made me think my algebra
was faulty. But I now realize there is a logical falsity in the
equations, at least in context. The context is betting on different
winners in a competition, using the most favourable odds available
from different bookies, such that winning one of three bets at
different odds (1/1, 2/1, 3/1) ensures that you cover the lost stakes
with the other two and still leave 10 units profit. Those equations
represented a hypothetical case, from which I hoped to devise a
general equation. But I can see now that there is a limit to how short
those odds can be without there being a logical impossibility
somewhere. Thanks for the help.

terpsichore334692 · Joined: 16 Aug 2007 Posts: 3

On Aug 12, 3:34 pm, Stan Brown wrote:
> Sun, 12 Aug 2007 13:05:48 +0100 from pmb :
>
> Don't post the same question multiple times. That just wastes
> everyone's time. I've already posted a response in alt.algebra.help.

The response on that forum from Frederick Williams was of some help,
as was the response here from Martin. Yours was of no help, either on
that forum or this one. So I don't know what you mean.

Frederick Williams · Joined: 05 Aug 2007 Posts: 20

terpsichore334692@yahoo.co.uk wrote:
>
> On Aug 12, 3:34 pm, Stan Brown wrote:
> > Sun, 12 Aug 2007 13:05:48 +0100 from pmb :
> >
> > Don't post the same question multiple times. That just wastes
> > everyone's time. I've already posted a response in alt.algebra.help.
>
> The response on that forum from Frederick Williams was of some help,
> as was the response here from Martin. Yours was of no help, either on
> that forum or this one. So I don't know what you mean.

He means cross-post (if you really must) don't multi-post. If someone
reads a cross-posted message in one group then it will appear as read in
all groups. This is not so with multi-posts, so someone who has read a
multi-post in one group may waste time and bandwidth by opening it in
another.

--
Remove "antispam" and ".invalid" for e-mail address.
"He that giveth to the poor lendeth to the Lord, and shall be repaid,"
said Mrs Fairchild, hastily slipping a shilling into the poor woman's
hand.

Martin · Joined: 05 Aug 2007 Posts: 22

wrote in message @l70g2000hse.googlegroups.com...
> On Aug 12, 2:14 pm, "Martin" wrote:
>> "pmb" wrote in message
>>
>> @bt.com...
>>
>> > It's a long time since I was at school, and I've forgotten a lot. But
>> > I'm
>> > trying to solve three equations. They look like they're solvable, but I
>> > can't do it. It's possible that they contain some absurdity that makes
>> > them unsolvable. But here they are:
>>
>> > a=10+b+c
>> > 2b=10+a+c
>> > 3c=10+a+b
>>
>> > Is there as solution?
>>
>> Yes. What have you done so far, and where did you get stuck?
>>
>> (Tip - take care to distinguish positive and negative values as you work
>> through it....)
>
> Well, it was getting a negative answer that made me think my algebra
> was faulty. But I now realize there is a logical falsity in the
> equations, at least in context. The context is betting on different
> winners in a competition, using the most favourable odds available
> from different bookies, such that winning one of three bets at
> different odds (1/1, 2/1, 3/1) ensures that you cover the lost stakes
> with the other two and still leave 10 units profit. Those equations
> represented a hypothetical case, from which I hoped to devise a
> general equation. But I can see now that there is a limit to how short
> those odds can be without there being a logical impossibility
> somewhere. Thanks for the help.
>

It seems you have proved mathematically (if such prooof were needed) that
the bookie always wins Smile

)

--
Martin
[ remove barrier to reply ]

terpsichore334692 · Joined: 16 Aug 2007 Posts: 3

On Aug 13, 5:32 pm, "Martin" wrote:
> wrote in message
>
> @l70g2000hse.googlegroups.com...
>
>
>
>
>
> > On Aug 12, 2:14 pm, "Martin" wrote:
> >> "pmb" wrote in message
>
> >>@bt.com...
>
> >> > It's a long time since I was at school, and I've forgotten a lot. But
> >> > I'm
> >> > trying to solve three equations. They look like they're solvable, but I
> >> > can't do it. It's possible that they contain some absurdity that makes
> >> > them unsolvable. But here they are:
>
> >> > a=10+b+c
> >> > 2b=10+a+c
> >> > 3c=10+a+b
>
> >> > Is there as solution?
>
> >> Yes. What have you done so far, and where did you get stuck?
>
> >> (Tip - take care to distinguish positive and negative values as you work
> >> through it....)
>
> > Well, it was getting a negative answer that made me think my algebra
> > was faulty. But I now realize there is a logical falsity in the
> > equations, at least in context. The context is betting on different
> > winners in a competition, using the most favourable odds available
> > from different bookies, such that winning one of three bets at
> > different odds (1/1, 2/1, 3/1) ensures that you cover the lost stakes
> > with the other two and still leave 10 units profit. Those equations
> > represented a hypothetical case, from which I hoped to devise a
> > general equation. But I can see now that there is a limit to how short
> > those odds can be without there being a logical impossibility
> > somewhere. Thanks for the help.
>
> It seems you have proved mathematically (if such prooof were needed) that
> the bookie always wins Smile

)
>

Well, no, all I've proved is that I'm not nearly as tuned in to this
stuff as I was ten years back. The frustrating thing is that in those
days I not only knew the mathematical logic, I also wrote a computer
program in Basic that let me punch in a series of bet odds, and the
target profit, and it would come back telling me what stakes to put on
each. The reason I didn't pursue it was that the profit margin was
usually such a small percentage of your outlay that it just wasn't
worth it. To make a respectable profit you needed stakes bigger than I
could afford. Also, with betting online, you had to open accounts with
at least a dozen bookies. But, to take an example, you could bet on a
football match, where there are three possible outcomes, Team A wins,
Team B wins, or it's a draw (tie). After studying the odds offered on
each outcome, you might take three different bets at three different
bookies, with a certainty of making a profit, whatever the result.
But, with football, or most sports, the profit margin is tiny. Where
you have a better chance of making a decent profit is with events like
Big Brother, where bookies' knowledge is no better than anyone else's,
and odds between bookies often vary greatly, especially near the start
of the event. But even now, with the UK Big Brother three quarters
over, you can still find the odds varying a lot. But the odds change
from hour to hour, so you have to be ready with a workable formula and
bang down your bets quick, before the odds change again.

Brian Reay · Joined: 05 Aug 2007 Posts: 215

"pmb" wrote in message @bt.com...
> It's a long time since I was at school, and I've forgotten a lot. But I'm
> trying to solve three equations. They look like they're solvable, but I
> can't do it. It's possible that they contain some absurdity that makes
> them unsolvable. But here they are:
>
> a=10+b+c
> 2b=10+a+c
> 3c=10+a+b
>
> Is there as solution?

There is no reason why a, b, and/or c could not be -ve.

There are several approachs but the golden rule is to have one equation per
unknown ie you need 1 equation to find 1 unknown, 2 for 2, etc.
Algebraic methods generally rely on this.

Here you've 3 of each so, in essence, if there is a solution you should be
able to find it. I'd start by reducing the problem to 2 unknowns, use the
first equation to subsitute for a in the other 2.

Therefore:

2b = 10 + 10+b+c +c or 2b = 20 + b + 2c or b = 20+2c

3c = 10 + 10+b+c +b or 3c = 20 + 2b +c or 2c= 20 + 2b or
c = 10 +b

You've now got 2 equations with 2 unknowns

b = 20+2c
c= 10+b

Repeat the process, reducing to 1 equ and 1 unknown

Use the first equ to subsitute in the second for b

c = 10 + 20 + 2c

-30 = c

Now back track:

b = 20 + 2c or b = 20 + 2(-30) so b = -40

You've now got b and c and use on of your original equations to find a

a = 10 + b + c so a = -60

So a = -60, b= -40, c = -30

Good practice is to check in all of your original equations, just in case:

(1) a=10+b+c -60 = 10 -40 - 30 OK
(2) 2b=10+a+c -80 = 10 -60- 30 OK
(3) 3c=10+a+b -90 = 10 -60 -40 OK

--
73
Brian, G8OSN
www.g8osn.org.uk

Now your amateur licence is free, why not send at least £15 per year to
support the
Radio Communications Foundation or STELAR?

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