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## #1 2013-03-10 02:01:49

BarandaMan
Guest

tinypic.com/view.php?pic=x1logw&s=6

There are 4 equations above, I do not know how many need to be used for this part of the question. But the question, as written in the image, is:

Derive an equation for a, that does not include a*, r or e.

## #2 2013-03-10 02:37:02

cmowla
Member
Registered: 2012-06-14
Posts: 62

It's really hard to read some of the exponents.  I think it might be better to rewrite the problem making sure that all symbols can be readable.  I "solved" a econ system of equations similar to this for a friend (but I think that was 11 equations), but I misunderstood the equations because he didn't explain it well enough to me.  So please rewrite the problem more legibly so that everyone is positive they are solving the correct problem.

In addition, are we to assume that all of these symbols represent variables (if so, is there any relationship between variables besides the given set of equations)?  Or are some of them constants?  If all are not variables, you need to specify because even if someone could have guessed correctly at what symbols you used, they will solve the wrong problem if they assume that some of these symbols are constants or not be able to solve the problem at all if indeed some symbols are variables and have a separate relationship with some or all of the other variables...and everyone loses.

Lastly, were you given any idea of what the form of the solution will be?  For example, are Taylor series expansions acceptable?

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## #3 2013-03-10 02:39:00

muxdemux
Member
Registered: 2012-12-23
Posts: 80

Solve the second equation for

, the third equation for
and the last equation for
. (Did I read your handwriting correctly?) Now you should have expressions for
,
and
.

into the the first equation (in your picture).

You will see that the resulting equation now contains

, so substitute your expression for
to eliminate it.

Finally, you should be left with an expression that contains

just like you did for
and
.

That should be it.

EDIT: I agree with cmowla. Your handwriting is difficult to read here. If any of your equations contain the letter t or number 3, then I've misread.

Last edited by muxdemux (2013-03-10 02:42:38)

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## #4 2013-03-10 03:04:30

BarandaMan
Guest

cmowla wrote:

It's really hard to read some of the exponents.  I think it might be better to rewrite the problem making sure that all symbols can be readable.  I "solved" a econ system of equations similar to this for a friend (but I think that was 11 equations), but I misunderstood the equations because he didn't explain it well enough to me.  So please rewrite the problem more legibly so that everyone is positive they are solving the correct problem.

In addition, are we to assume that all of these symbols represent variables (if so, is there any relationship between variables besides the given set of equations)?  Or are some of them constants?  If all are not variables, you need to specify because even if someone could have guessed correctly at what symbols you used, they will solve the wrong problem if they assume that some of these symbols are constants or not be able to solve the problem at all if indeed some symbols are variables and have a separate relationship with some or all of the other variables...and everyone loses.

Lastly, were you given any idea of what the form of the solution will be?  For example, are Taylor series expansions acceptable?

Firstly, I hope this picture is more clear: i49.tinypic.com/r8iic2.jpg

There are 4 parameters. Alpha, Beta, Yamma and the Circle with the line through.
Then a, a*, b, b*, g, g*, r and e.

These are meanings behind the letters, but there are no other links between these variables apart from those in the equations.

The *'d variables represent the variable a, b and g but just for a country abroad. The very first question is to derive an expression for a, that does not contain a*, e, or r. This is then used in the next 6 questions to continue the question.

I hope this clears some things up. And muxdemux, I will try this now. I have done so much rearranging for variables and I keep essentially getting: a=a / 0=0 i.e. complete equality...not an expression.

And cmowla, we have not yet covered Taylor expansions so I think it is definitely a case of rearranging/substituting.

Really look forward to hearing from you.

## #5 2013-03-10 03:22:44

BarandaMan
Guest

Ok, I really look forward to replies on this. But I have got something which MAY or (very may well not be) correct, after following muxdemux's reply:

tinypic.com/view.php?pic=23jr5ox&s=6

Do any of you get the same? :D:D:D

## #6 2013-03-10 04:33:47

cmowla
Member
Registered: 2012-06-14
Posts: 62

I solved for A, elliminating A*,e,r, treating all symbols as variables.  Note that I capitalized a to A to differentiate it between alpha a little more.  In fact, you will find from the work below that alpha does not even make up A at all.

I provided all steps I took so that you can double check my work (I don't solve systems of equations that often anymore).

Problem:
Solve for A in terms of anything but A*,e, and r using the following system of equations:

Solution:

EDIT: I corrected the mistake and got the correct answer.

Last edited by cmowla (2013-03-10 05:50:53)

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## #7 2013-03-10 04:48:02

bob bundy
Registered: 2010-06-20
Posts: 8,065

hi BarandaMan

What you have in post 5 is algebraically correct, but you would have to bring all the a terms to one side.

Here's my method:

eqn.1 + 2 to eliminate e

3 - 4 to eliminate r

Now you can add these to eliminate a-star

Your version will become this with a bit of algebraic manipulation.

Just checking if cmowla's comes to the same thing ......

Bob

Last edited by bob bundy (2013-03-10 04:50:38)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #8 2013-03-10 05:05:18

bob bundy
Registered: 2010-06-20
Posts: 8,065

hi cmowla

Sorry.  I'm not getting the same as you.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #9 2013-03-10 05:13:03

cmowla
Member
Registered: 2012-06-14
Posts: 62

bob bundy wrote:

hi cmowla

Sorry.  I'm not getting the same as you.

Bob

Is there a mistake in my work?  I provided all steps.

EDIT:
I found the mistake.  It's in the last step before "step 3".  I'll fix it and see if I got the same answer as you.

Last edited by cmowla (2013-03-10 05:20:04)

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## #10 2013-03-10 05:23:33

BarandaMan
Guest

Really appreciate the replies; but obviously a tad confused because two people have two different answers

On cmowla's post, why does the denominator (1-theta) just change to -(theta-1) ??

## #11 2013-03-10 05:44:10

cmowla
Member
Registered: 2012-06-14
Posts: 62

BarandaMan wrote:

Really appreciate the replies; but obviously a tad confused because two people have two different answers

I corrected a mistake in my work (I wrote a beta in the denominator instead of a gamma) when I substituted equ3b into equ4b, and I now have the same answer as bob bundy.

BarandaMan wrote:

On cmowla's post, why does the denominator (1-theta) just change to -(theta-1) ??

Because they are equal (just multiply in that negative one).  This is useful to do sometimes for algebraic simplification.

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## #12 2013-03-10 05:48:05

BarandaMan
Guest

This is fantastic! Both your answers are equivalent, I went through them both! This all makes sense!

Thank you both so much!

The follow up question seems harder, it is using all the same 4 equations, but derive an expression for 'e', without using 'e','r', or 'a*'....

How does one generally go about something like this, when it is so easy to eliminate the 'e's' in equations 1 and 2? How can we keep the e?

## #13 2013-03-10 05:48:54

BarandaMan
Guest

cmowla wrote:
BarandaMan wrote:

Really appreciate the replies; but obviously a tad confused because two people have two different answers

I corrected a mistake in my work (I wrote a beta in the denominator instead of a gamma) when I substituted equ3b into equ4b, and I now have the same answer as bob bundy.

BarandaMan wrote:

On cmowla's post, why does the denominator (1-theta) just change to -(theta-1) ??

Because they are equal (just multiply in that negative one).  This is useful to do sometimes for algebraic simplification.

They are equal, that was very useful for eliminating A, so we could make it 2A on the LHS, really thank you for the trick.

## #14 2013-03-10 05:55:09

cmowla
Member
Registered: 2012-06-14
Posts: 62

BarandaMan wrote:

They are equal, that was very useful for eliminating A, so we could make it 2A on the LHS, really thank you for the trick.

You're welcome.  Go look at the post again.  I realized I used the wrong code for phi, but I corrected it now.

Last edited by cmowla (2013-03-10 05:55:41)

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## #15 2013-03-10 05:56:56

BarandaMan
Guest

Despite my last post, I think it is actually easier to derive an equation for e?

I get,

e=A - (theta(b*+gamma(A)-b))/gamma - g.

From equation (1),re-wrote it in terms of 'e'.

Then from equation (4), re-wrote that in terms of A*, and put that A* term into the one above.

Then from equation (3), re-wrote that in terms of r, and put that r term into the one above.

And it gives: e=A - (theta(b*+gamma(A)-b))/gamma - g.

Does everyone else get this or is this wrong?
Is there any way to get better and this kind of algebra? Practise makes perfect? I can't just see the tricks though

## #16 2013-03-10 05:58:14

BarandaMan
Guest

cmowla wrote:
BarandaMan wrote:

They are equal, that was very useful for eliminating A, so we could make it 2A on the LHS, really thank you for the trick.

You're welcome.  Go look at the post again.  I realized I used the wrong code for phi, but I corrected it now.

Thank you so much, I have written up your post so neatly and it really has helped me break down such solutions step by step which is exactly what I have trouble doing. Really appreciate your time

## #17 2013-03-10 06:21:28

bob bundy
Registered: 2010-06-20
Posts: 8,065

hi BarandaMan,

e=A - (theta(b*+gamma(A)-b))/gamma - g

Where did theta come from?

and did you mean

So I have checked this

and it doesn't work out when I substitute numbers.

Explain your steps and I'll try to spot where you are going wrong.

Bob

Last edited by bob bundy (2013-03-10 06:40:30)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #18 2013-03-10 06:32:10

BarandaMan
Guest

bob bundy wrote:

hi BarandaMan,

e=A - (theta(b*+gamma(A)-b))/gamma - g

Where did theta come from?

and did you mean

So I have checked this

and it doesn't work out when I substitute numbers.

I'll have a go and see what I get.

Bob

Yes sorry, I meant phi, I need to revise the Greek alphabet

But I got :

as you posted, but the middle whole term divided by gamma

Really appreciate the help. Your method for the a expression is brilliantly intuitive and I have written that out in full. Thank you.

## #19 2013-03-10 06:46:45

bob bundy
Registered: 2010-06-20
Posts: 8,065

hi

If you mean this:

then it works with numbers, so I think it is ok.  I got a different expression because I kept in alpha g and gstar but not gamma and b.

Bob

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #20 2013-03-10 07:03:58

bob bundy
Registered: 2010-06-20
Posts: 8,065

Oh yes.  You might find the following helpful in checking your algebra.

I used a spreadsheet to substitute numbers into the equations.

First I entered the 12 variable names in a row.

Then I chose numbers for those that occur in equation 1 and made a formula to calculate 'a'.

Then I constructed the equation for astar.  Of course I'd already fixed the value of astar but I hadn't chosen the value for gstar yet.  So I just fiddled the value of gstar so that astar came out correctly.

Then I made the equations for b and bstar.

So that gave me 12 variables with values consistent with the 4 equations.

When I had finished my equation for a, I evaluated it using that formula and my chosen variables.  You can see in my screen shot below that it evaluated to 51 as required.  And I checked yours and it too gave 51.

Now that trick is not 100% foolproof.  You could have errors that 'cancel out' leaving an answer that looks correct when it isn't.  But it's very unlikely with those numbers, so I was pretty confident we were both using correct formulas.

I saved the spreadsheet, so when you offered an equation for 'e', all I had to do was construct the equation in the spreadsheet and look to see if e came out correctly.. It did!

Bob

Last edited by bob bundy (2013-03-10 07:08:09)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #21 2013-03-10 07:10:15

BarandaMan
Guest

bob bundy wrote:

Oh yes.  You might find the following helpful in checking your algebra.

I used a spreadsheet to substitute numbers into the equations.

First I entered the 12 variable names in a row.

Then I chose numbers for those that occur in equation 1 and made a formula to calculate 'a'.

Then I constructed the equation for astar.  Of course I'd already fixed the value but I hadn't chosen a value for gstar yet.  So I just fiddled the value of gstar so that astar came out correctly.

Then I made the equations for b and bstar.

So that gave me 12 variables with values consistent with the 4 equations.

When I had finished my equation for a, I evaluated it using that formula and my chosen variables.  You can see in my screen shot below that it evaluated to 51 as required.  And I checked yours and it too gave 51.

Now that trick is not 100% foolproof.  You could have errors that 'cancel out' leaving an answer that looks correct when it isn't.  But it's very unlikely with those numbers, so I was pretty confident we were both using correct formulas.

I saved the spreadsheet, so when you offered an equation for 'e', all I had to do was construct the equation in the spreadsheet and look to see if e came out correctly.. It did!

Bob

This is absolutely incredible. Thank you so much. I should now be able to tackle the third part of the question. This has all been really helpful.

I will try not to derive the expression for 'e' with alpha g and g* to see if I can do it, I love this kind of substitution when it goes correct!

Thank you! When I have an expression with g* and g, I will post it here and if you could maybe let me know if it is what you had, that would be great! Please don't post it before hand, I really want to see if i can do this hahaha!
Thank you Bob!!!

## #22 2013-03-10 07:12:26

bob bundy
Registered: 2010-06-20
Posts: 8,065

You're welcome.

Bob

ps. the spreadsheet trick 'works' because the rules of algebra are just the rules of arithmetic.  So, if it doesn't work with numbers, then the algebra must be wrong.

Last edited by bob bundy (2013-03-10 07:13:59)

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei

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## #23 2013-03-10 07:15:16

BarandaMan
Guest

That excel trick is incredible. It actually holds for all 4 equations and gave 51 for my answer as you said, that is great that we can check with real numbers. I willt try it with new expression! Thank you!

## #24 2013-03-10 07:27:42

BarandaMan
Guest

bob bundy wrote:

hi

If you mean this:

then it works with numbers, so I think it is ok.  I got a different expression because I kept in alpha g and gstar but not gamma and b.

Bob

Oh no, for 'a' i am getting the trick correct in excel.

But with 'e' as above, the numbers aren't equalling e=3 as they should? Do you get this too?

BarandaMan
Guest