Help with the most disgusting 'solve for x' question I have been given

Srs

Unparseable latex formula:

[br][br]\displaystyle \frac{A-x}{A+x} = \frac{A-y}{A+y} \frac{A-z}{A+z}[br][br][br]\[\textrm{The answer is:}\][br] [br]x = \displaystyle \frac{y+z}{1+ (\frac{yz}{A^2})}[br][br]

ffs didn't mean to post. was trying to edit my CORRECT latex code, but the code here is messed up, some help on the fractions would be sweet.

(edited 8 years ago)

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Reply 1

Zacken

Original post by Et Tu, Brute?

Srs

\displaystyle \frac{A-x}{A+x} = \frac{A-y}{A+y} \frac{A-z}{A+z}

The answer is:

x = \frac{}{}

Assuming that what I've written is what you meant, then let's call everyting on the RHS

k

Unparseable latex formula:

\displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

Now collecting the x terms:

\displaystyle A - kA = kx + x

.

Factorise, simplify, isolate x and back-substitute

k

Reply 2

Et Tu, Brute?

Original post by Zacken

Assuming that what I've written is what you meant, then let's call everyting on the RHS

k

Unparseable latex formula:

\displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

Now collecting the x terms:

\displaystyle A - kA = kx + x

.

Factorise, simplify, isolate x and back-substitute

k

thanks, that's what I meant, what is the code you used?

Reply 3

Zacken

Original post by Et Tu, Brute?

thanks, that's what I meant, what is the code you used?

No problem. You should be able to see the code in my reply, but that's of no import right now. Do you understand the method I'm showcasing? Can you follow it?

Reply 4

TeeEm

Original post by Zacken

... showcasing ...

very nice!

Reply 5

Et Tu, Brute?

Original post by Zacken

Assuming that what I've written is what you meant, then let's call everyting on the RHS

k

Unparseable latex formula:

\displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

Now collecting the x terms:

\displaystyle A - kA = kx + x

.

Factorise, simplify, isolate x and back-substitute

k

Original post by Zacken

No problem. You should be able to see the code in my reply, but that's of no import right now. Do you understand the method I'm showcasing? Can you follow it?

you edited in my post, so I can't actually see that part, only the part in your own post which is like:

\frac{A-x}{A+x}

which is what I have in the op

Yeah makes sense, though it doesn't look like that will end up like the answer I have, which brings it back to the latex code, how do I do it? The answer is much messier than the starting expression.

Reply 6

Zacken

Original post by Et Tu, Brute?

Just click on the equation and it'll bring up the code.

You'll need to clean up your answer. It'll come down to simplifying

\frac{1-k}{1+k}

Reply 7

Et Tu, Brute?

Original post by Zacken

Just click on the equation and it'll bring up the code.

You'll need to clean up your answer. It'll come down to simplifying

\frac{1-k}{1+k}

op is edited.

Now I have that out of the way I'll try it out again, but can't see it bringing me to the answer from

\frac{1-k}{1+k}

, maybe I'm just being pessimistic, its been a long day.

Reply 8

Zacken

Assuming that what I've written is what you meant, then let's call everyting on the RHS

k

Unparseable latex formula:

\displaysyle \frac{A-x}{A+x} = k \Rightarrow A - x = k(A+x) \Rightarrow A - x = kA + kx

Now collecting the x terms:

\displaystyle A - kA = kx + x

.

Factorise, simplify, isolate x and back-substitute

k

Original post by Et Tu, Brute?

op is edited.

Now I have that out of the way I'll try it out again, but can't see it bringing me to the answer from

\frac{1-k}{1+k}

, maybe I'm just being pessimistic, its been a long day.

Can you understand the method in my post above? It simplifies to

A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}

Reply 9

Et Tu, Brute?

Original post by Zacken

Can you understand the method in my post above? It simplifies to

A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}

yeah makes sense,working on it now. Will update

Reply 10

Et Tu, Brute?

Original post by Zacken

Can you understand the method in my post above? It simplifies to

A(1-k) = x(k+1) \Rightarrow x = \frac{A(1-k)}{1+k}

I end up with a complete abomination of an expression

Reply 11

Zacken

Original post by Et Tu, Brute?

I end up with a complete abomination of an expression

You'll need to look for common factors and cancel them out.

Reply 12

Et Tu, Brute?

Original post by Zacken

You'll need to look for common factors and cancel them out.

have you worked this through to the answer in the op?

I am more than happy to spend all night at this, however right not I am not convinced I will end up with the answer in the op, which is killing my motivation to actually do this tedious elementary maths

Reply 13

Zacken

Original post by Et Tu, Brute?

Nopes. I can see off the bat that a few things cancel, but I'm not convinced of the given answer myself.

Reply 14

Et Tu, Brute?

Original post by Zacken

Nopes. I can see off the bat that a few things cancel, but I'm not convinced of the given answer myself.

****ing knew it.

This is why I hate physics.

Reply 15

Zacken

Original post by Et Tu, Brute?

****ing knew it.

This is why I hate physics.

I'm not saying it's incorrect, just that I haven't verified it myself. I'll do that in a bit and tell you.

Reply 16

Et Tu, Brute?

Original post by Zacken

I'm not saying it's incorrect, just that I haven't verified it myself. I'll do that in a bit and tell you.

wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.

Reply 17

Zacken

Original post by Et Tu, Brute?

wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.

Oh well, you can still solve it and get a correct answer that isn't the given one.

Reply 18

Et Tu, Brute?

Original post by Zacken

Oh well, you can still solve it and get a correct answer that isn't the given one.

well it seemed odd to me that we were asked to rearrange an equation in a physics modules tbh, it makes sense now that it isn't a simple but tedious maths exercise. It most likely involves the inverse Lorentz boost, where exactly it comes in is a different matter altogether. But I suspect that solving that equation for x and taking it no further will yield little if any marks sadly.

Reply 19

Gregorius

Original post by Et Tu, Brute?

wolfram alpha'd it. Equations don't equal each other according to that, so I wouldn't waste your time. There is probably some substitution to be made somewhere.