What is the Transposition Problem for Algebra?

[FaraDazed]

Homework Statement

Transpose:

${D \over d} = \sqrt{\frac{f + p}{f - p}}$

To make f the subject...

The Attempt at a Solution

This is neither homework or coursework, it was a little challenge set by my math tutor.

To be honest I am lost from the start, can someone just let me know if I am on the right track...

$({D \over d})^2 = {f + p \over f - p}$

 

[swsw]

Seems right, manipulate it further and you will get "f" as subject.

 

[Chestermiller]

Here is an interesting trick you can use. If each side of an equation is a fraction, the numerator plus the denominator divided by the numerator minus the denominator of one side of the equation is equal to the numerator plus the denominator divided by the numerator minus the denominator of the other side of the equation.

 

[FaraDazed]

OK well I am am getting no where to be honest. Could I use the above rule to help me in this problem?

I don't know if the below is going to help or if i were more on the right tracks before.

${D \over d} = \sqrt{\frac{f + p}{f - p}}\\ D(f-p)^2= d\sqrt{f+p}\\$

 

[SammyS]

OK well I am am getting no where to be honest. Could I use the above rule to help me in this problem?

I don't know if the below is going to help or if i were more on the right tracks before.

${D \over d} = \sqrt{\frac{f + p}{f - p}}\\ D(f-p)^2= d\sqrt{f+p}\\$

No.

That doesn't work at all.

Homework Statement

Transpose:

${D \over d} = \sqrt{\frac{f + p}{f - p}}$

To make f the subject...

The Attempt at a Solution

This is neither homework or coursework, it was a little challenge set by my math tutor.

To be honest I am lost from the start, can someone just let me know if I am on the right track...

$({D \over d})^2 = {f + p \over f - p}$

That's a good first step.

Now, multiply both sides by d²(f-p). You get the same result by "cross multiplying".

Now distribute the D² and d².

Continue on ...

 

[FaraDazed]

^^OK thanks for you help its much appreciated...
So after multiplying both sides as you said, then cancelling out the bits that can be cancelled, am I left with the following...

$(f-p)D^2 = (f+p)d^2$

If so, the only thing, apart from isolating one of those expressions is to do the following (if its correct)

$fD^2 - pD^2 = fd^2 + pd^2$

then could I do...

$fD^2 = pD^2 + fd^2 + pd^2 \\ fD^2 - fd^2 = pD^2 + pd^2 \\ fD^2 = {d^2 + pD^2 + pd^2 \over f} \\ (fD^2)f = d^2 + pD^2 + pd^2$

?

Also a question I have; as the problem asks to make 'f' the subject, would f²=... be ok or does it have to be f = ... ?

 

[FaraDazed]

I have kept going on the basis the above is right (probably isn't) but anyway...

$(fD^2)f - pD^2 = d^2 + pd^2 \\ (fD^2)f - D^2 = {d^2 + pd^2 \over p } \\ (fD^2)f - D^2 = d^2 + d^2 \\ (fD^2)f - D^2 = d^4 \\ f^2D^2 - D^2 = d^4 \\ f^2 = d^4$

 

[FaraDazed]

EDIT:

OK thanks very much.

so that's the easy way then?

Does the way I done it make sense though?

 

[Vineeth T]

Yeah!
But as you said what I have done is the easiest method.

 

[FaraDazed]

Yeah!
But as you said what I have done is the easiest method.

the last line is it "- p" or should it just be "p" next to the fraction as if it were times by?

 

[Vineeth T]

Yes! It is that fraction times p.
I have edited it.

 

[HallsofIvy]

^^OK thanks for you help its much appreciated...



So after multiplying both sides as you said, then cancelling out the bits that can be cancelled, am I left with the following...

$(f-p)D^2 = (f+p)d^2$

If so, the only thing, apart from isolating one of those expressions is to do the following (if its correct)

$fD^2 - pD^2 = fd^2 + pd^2$

then could I do...

$fD^2 = pD^2 + fd^2 + pd^2 \\ fD^2 - fd^2 = pD^2 + pd^2$

You are good up to here.

$fD^2 = {d^2 + pD^2 + pd^2 \over f}$

I can't imagine how you think this is the same! You have added the $d^2$ and divided by f. You can't do that with "$fd^2$". Since you want to solve for f and already have every instance of "f" on the right side, just factor out f: $f(D^2- d^2)=$

$(fD^2)f = d^2 + pD^2 + pd^2$

?

Also a question I have; as the problem asks to make 'f' the subject, would f²=... be ok or does it have to be f = ... ?

 

[FaraDazed]

You are good up to here. I can't imagine how you think this is the same! You have added the $d^2$ and divided by f. You can't do that with "$fd^2$". [/itex]

I can see now why I can do that, I don't know why I thought I could either!

Sorry for bumping this back up but can you check my new solution pleas? I have completely started from fresh and used the componendo et dividendo rule to make it a lot quicker and easier. ${D \over d } = \sqrt{f+p \over f-p} \\ {D^2 \over d^2} = {f+p \over f-p} \\ {D^2 + d^2 \over D^2 - d^2} = {f \over p } \\ f = ({\frac {D^2 + d^2}{D^2 + d^2}}) p \\$

There were go, could anyone check this for me pleas?

 

[Mark44]

I can see now why I can do that, I don't know why I thought I could either!

Sorry for bumping this back up but can you check my new solution pleas? I have completely started from fresh and used the componendo et dividendo rule to make it a lot quicker and easier. ${D \over d } = \sqrt{f+p \over f-p} \\ {D^2 \over d^2} = {f+p \over f-p} \\ {D^2 + d^2 \over D^2 - d^2} = {f \over p } \\$

Above is correct; you have a mistake below.

$f = ({\frac {D^2 + d^2}{D^2 + d^2}}) p \\$

There were go, could anyone check this for me pleas?

 

[FaraDazed]

Above is correct; you have a mistake below.

Is it the way it is bracketed and times by p or is it that now it has moved over it should be the sqaure root of it all times by p?

EDIT: Just remembered this solution had already been given by Vieeneth T

 

[Mark44]

Is it the way it is bracketed and times by p or is it that now it has moved over it should be the sqaure root of it all times by p?

No, that has nothing to do with it. Start with where you were correct, and work through the steps for solving for f. It's pretty simple.

EDIT: Just remembered this solution had already been given by Vieeneth T

Since giving complete solutions isn't permitted at this forum, his post has been deleted.

 

[FaraDazed]

OK I think it was just a typo, I put a + when it should have been a -, if that is the error that was just a typo :)

$f = ({\frac {D^2 + d^2}{D^2 - d^2}}) p \\$

 

[Mark44]

OK I think it was just a typo, I put a + when it should have been a -, if that is the error that was just a typo :)

$f = ({\frac {D^2 + d^2}{D^2 - d^2}}) p \\$

That's better!