# Changing the subject of a formula

• Aug 8th 2010, 08:21 PM
PythagorasNeophyte
Changing the subject of a formula
A very good day everyone. I am urgently requesting for your help for this taxing mathematics questions, all relating to algebra (changing the subject of a formula).
If you could help me to explain those sums I would be very gratified. So here they are:

In each of the following cases, rewrite the equation such that the letter in the square brackets is the subject of a given formula.

1. $\displaystyle 1/u + 1/v =1/f$ [u]

2. $\displaystyle y = 3x + 5/2-z$ [z]

3. $\displaystyle T = 2\pi \sqrt{L/g}$ [g]

4. $\displaystyle s = v^2 - u^2 / 2a$ [u]

5. $\displaystyle y = t^2 - x^2 / t^2 + x^2$ [x]
• Aug 8th 2010, 09:00 PM
eumyang
You may want to start using the \frac command in LaTeX to input fractions properly. For example \frac{a}{b} would give you this: $\displaystyle \frac{a}{b}$. I say this because I have a feeling there will be some confusion as to what you wrote. For example, #2, is it this:
$\displaystyle y = 3x + \frac{5}{2}-z$
or this?
$\displaystyle y = \dfrac{3x + 5}{2-z}$
#4 and #5 presents the same problems on interpretation.

#3 looks straightforward, so I'll do that.
$\displaystyle T = 2\pi \sqrt{\frac{L}{g}}$

Divide both sides by 2 pi:
$\displaystyle \dfrac{T}{2\pi} = \sqrt{\dfrac{L}{g}}$

Square both sides:
$\displaystyle \dfrac{T^2}{4\pi^2} = \dfrac{L}{g}$

Cross-multiply:
$\displaystyle gT^2 = 4\pi^2 L$

And divide both sides by T^2:
$\displaystyle g = \dfrac{4\pi^2 L}{T^2}$
• Aug 8th 2010, 11:01 PM
PythagorasNeophyte
Reformatting...
I am very sorry, eumyang. I haven't been acquainted with LaTeX recently. So here are the questions again-

1. $\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ [v]

2. $\displaystyle y = 3x + \frac{5}{2 -x}$ [z]

3. $\displaystyle y = \frac{t^2 - x^2}{t^2 + x^2}$ [x]

4. $\displaystyle s = \frac{v^2 - u^2}{2a}$ [u]

Anyway, thank you so much for your constructive explanation on one of the questions. But I would appreciate if all of my questions are done. Thank you so much!
• Aug 8th 2010, 11:20 PM
CaptainBlack
Quote:

Originally Posted by PythagorasNeophyte
I am very sorry, eumyang. I haven't been acquainted with LaTeX recently. So here are the questions again-

1. $\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ [v]

2. $\displaystyle y = 3x + \frac{5}{2 -x}$ [z]

3. $\displaystyle y = \frac{t^2 - x^2}{t^2 + x^2}$ [x]

4. $\displaystyle s = \frac{v^2 - u^2}{2a}$ [u]

Anyway, thank you so much for your constructive explanation on one of the questions. But I would appreciate if all of my questions are done. Thank you so much!

This is not a homework service.

What have you done, and what are the difficulties you are having?

CB
• Aug 9th 2010, 01:36 AM
PythagorasNeophyte
Dear CB,

Here's what I have done:

1. $\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$

$\displaystyle \frac{fv}{fuv} + \frac{fu}{fuv} = \frac{uv}{fuv}$

$\displaystyle \frac{fv + fu}{fuv} = \frac{uv}{fuv}$

$\displaystyle fuv(fv + fu) = fuv(uv)$

$\displaystyle f^2uv^2 + f^2u^2v = fu^2v^2$ ???

And then I'm stuck.

---------------

2. $\displaystyle y = 3x + \frac{5}{2 -x}$

$\displaystyle \frac{5}{2 -x} = y - 3x$

$\displaystyle 2 - x = \frac{y - 3x}{5}$

$\displaystyle x = \frac{y - 3x}{5} - 2$ ???

--------------

3. $\displaystyle y = \frac{t^2 - x^2}{t^2 + x^2}$

$\displaystyle \frac{y}{1} = \frac{t^2 - x^2}{t^2 + x^2}$

$\displaystyle yt^2 + yx^2 = t^2 - x^2$
• Aug 9th 2010, 02:17 AM
CaptainBlack
Quote:

Originally Posted by PythagorasNeophyte
Dear CB,

Here's what I have done:

1. $\displaystyle \dfrac{1}{u} + \dfrac{1}{v} = \dfrac{1}{f}$

so:

$\displaystyle \dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{d}=\dfrac{d-f}{df}$

so:

$\displaystyle v=\dfrac{df}{d-f}$

CB
• Aug 9th 2010, 02:20 AM
CaptainBlack
Quote:

Originally Posted by PythagorasNeophyte
I am very sorry, eumyang. I haven't been acquainted with LaTeX recently. So here are the questions again-

1. $\displaystyle \frac{1}{u} + \frac{1}{v} = \frac{1}{f}$ [v]

2. $\displaystyle y = 3x + \frac{5}{2 -x}$ [z]

3. $\displaystyle y = \frac{t^2 - x^2}{t^2 + x^2}$ [x]

4. $\displaystyle s = \frac{v^2 - u^2}{2a}$ [u]

Anyway, thank you so much for your constructive explanation on one of the questions. But I would appreciate if all of my questions are done. Thank you so much!

there is a typo in 2. You seem to be asked to make z the subject but z does not occur in the equation.

CB
• Aug 9th 2010, 02:23 AM
CaptainBlack
Quote:

Originally Posted by PythagorasNeophyte
3. $\displaystyle y = \frac{t^2 - x^2}{t^2 + x^2}$

$\displaystyle \frac{y}{1} = \frac{t^2 - x^2}{t^2 + x^2}$

$\displaystyle yt^2 + yx^2 = t^2 - x^2$

collect the $\displaystyle x^2$ terms:

$\displaystyle (y +1)x^2=(1-y)t^2$

$\displaystyle x^2=\dfrac{1-y}{1+y}\ t^2$

etc

CB
• Aug 10th 2010, 12:13 AM
PythagorasNeophyte
Thank you so much for your help, CaptainBlack.

Indeed, there is a typo error in question 2, where the $\displaystyle x$ I put before should be replaced with$\displaystyle z.$ Hence, here is my reworking of question 2:

$\displaystyle y = 3x + \frac{5}{2 -z}$

And this is my working:

$\displaystyle y = 3x + \frac{5}{2 -z}$

$\displaystyle \frac{5}{2 -x} = y - 3z$

$\displaystyle 2 - x = \frac{y - 3z}{5}$

$\displaystyle x = \frac{y - 3x}{5} - 2$

$\displaystyle x= \frac{y - 3x - 10}{5}$
---------------

This is my working for question 4:

$\displaystyle s = \frac{v^2 - u^2}{2a}$

$\displaystyle v^2 - u^2 = 2aS$

$\displaystyle -u^2 = 2aS - v^2$

$\displaystyle u = +-\sqrt2aS - v^2$
• Aug 10th 2010, 02:16 AM
HallsofIvy
Quote:

Originally Posted by PythagorasNeophyte
Thank you so much for your help, CaptainBlack.

Indeed, there is a typo error in question 2, where the $\displaystyle x$ I put before should be replaced with$\displaystyle z.$ Hence, here is my reworking of question 2:

$\displaystyle y = 3x + \frac{5}{2 -z}$

And this is my working:

$\displaystyle y = 3x + \frac{5}{2 -z}$

$\displaystyle \frac{5}{2 -x} = y - 3z$

??? You simply swapped 'x' and 'z'. That's not a legal algebraic step.
From $\displaystyle y= 3x+ \frac{5}{2- z}$ you get $\displaystyle \frac{5}{2- z}= y- 3x$, by subtracting 3x from both sides.
Now invert both sides getting $\displaystyle \frac{2- z}{5}= y- 3x$.

From [tex]\frac{2- z}{5}= y- 3x, multiply both sides by 5 to get 2- z= 5(y- 3x)= 5y- 15x. Now subtract 2 from both sides: -z= 5y- 15x- 2, and then multiply both sides by -1 to get z= -5y+ 15x+ 2.

Quote:

$\displaystyle 2 - x = \frac{y - 3z}{5}$

$\displaystyle x = \frac{y - 3x}{5} - 2$

$\displaystyle x= \frac{y - 3x - 10}{5}$
---------------

This is my working for question 4:

$\displaystyle s = \frac{v^2 - u^2}{2a}$

$\displaystyle v^2 - u^2 = 2aS$

$\displaystyle -u^2 = 2aS - v^2$

$\displaystyle u = +-\sqrt2aS - v^2$
Last step is wrong. $\displaystyle -u^2$ will never be positive so you simply can't do this. Also you cannot just take the square root of part of the right side. What ever you do to one side of an equation, you must do to the entire side.

From $\displaystyle -u^2= 2aS- v^2$, multiply both sides by -1 to get $\displaystyle u^2= v^2- 2aS$. NOW you can take the square root of both sides: $\displaystyle u= \pm\sqrt{v^2- 2aS}$.
• Aug 10th 2010, 03:01 AM
PythagorasNeophyte
Queries
Thank you so much, HallsofIvy, for taking your time.

In the first one, I can't interpret how you had changed $\displaystyle \frac{5}{2- z}= y- 3x$ into $\displaystyle \frac{2- z}{5}= y- 3x$. I am not sure why
$\displaystyle \frac{5}{2- z}$ can be inverted like this, viz. $\displaystyle \frac{3}{4}$ cannot be changed into $\displaystyle \frac{4}{3}$ without any calculations.
---------------

In the second one, you had stated when multiplying $\displaystyle -1$ to $\displaystyle -u^2= 2aS- v^2$ at both sides gives $\displaystyle u^2= v^2- 2aS$. Why is the $\displaystyle v^2$ placed before $\displaystyle -2aS$?
Can $\displaystyle u^2 = -2aS + v^2$ be accepted?

Thank you for your time clarifying. Hope to hear from you soon.
• Aug 10th 2010, 03:52 AM
eumyang
Quote:

Originally Posted by PythagorasNeophyte
In the first one, I can't interpret how you had changed $\displaystyle \frac{5}{2- z}= y- 3x$ into $\displaystyle \frac{2- z}{5}= y- 3x$. I am not sure why
$\displaystyle \frac{5}{2- z}$ can be inverted like this, viz. $\displaystyle \frac{3}{4}$ cannot be changed into $\displaystyle \frac{4}{3}$ without any calculations.

Looks like a typo on his part.

After you "flip" both sides, multiply both sides by 5, then by -1 (or combine by multiplying both sides by -5) and then add 2 to both sides:
\displaystyle \begin{aligned} y &= 3x + \dfrac{5}{2 -z} \\ y - 3x &= \dfrac{5}{2 -z} \\ \dfrac{1}{y - 3x} &= \dfrac{2 -z}{5} \\ \dfrac{5}{y - 3x} &= 2 -z \\ -\dfrac{5}{y - 3x} &= z - 2 \\ z &= 2 -\dfrac{5}{y - 3x} \\ \end{aligned}