MHB Rearranging Formulas: Finding W in Terms of M and L

[Samwise-zambeezi]

Hi,

I have a formula which I need to rearrange but it's been a while since I covered this at school, I've tried to read up and re-study but I keep getting confused about what I can/can't do to the various brackets.

The formula is:

M = (((W x L / 2) x (L / 2)) - ((W x L / 2) x (L / 4)))

I'm trying to rearrange in terms of W, so W = ...

Any help or advance anyone could offer would be greatly appreciated.

Thanks

 

[Jameson]

Hi! Can you start by multiplying each term in this by 4? What do you get if you do that?

Is this the starting formula by the way?

$$M = \left( \frac{W \times L}{2} \times \frac{L}{2} \right) - \left( \frac{W \times L}{2} \times \frac{L}{4} \right)$$

 

[Samwise-zambeezi]

Hi Jameson, Happy New Year and thanks very much for getting back to me!

So, the equation definitely looks better in that format (i was using microsoft excel before). I'll try to mutiply both terms by 4 below (Sorry, I'm not used to using proper math input formatting)...

\[ M=(\frac{W x L}{2}) x (\frac{L}{2}) - (\frac{W x L}{2}) x (\frac{L}{4}) \]

\[ Mx4=(\frac{W x L}{2}) x (\frac{L}{2}) x 4 - (\frac{W x L}{2}) x (\frac{L}{4})x 4 \]

\[ Mx4=(({W x Lx2}) x (Lx2)) - ({W x Lx2}) x L \]

Then, if that's right?... could i divide everything by 'L'?

\[ (\frac{Mx4}{L})=(({W x Lx2}) x L) - ({W x Lx2}) \]

I'm not really sure about that, or if it's right, where to go from here. Think I'm getting confused about which parts of the term i should apply the 'action' to (ie. both sides of the '-', or both sides of the 'x' on both sides of the '-')

Sorry, hard to explain in writing but really appreciate your help

 

[topsquark]

Hi Jameson, Happy New Year and thanks very much for getting back to me!

So, the equation definitely looks better in that format (i was using microsoft excel before). I'll try to mutiply both terms by 4 below (Sorry, I'm not used to using proper math input formatting)...

\[ M=(\frac{W x L}{2}) x (\frac{L}{2}) - (\frac{W x L}{2}) x (\frac{L}{4}) \]

\[ Mx4=(\frac{W x L}{2}) x (\frac{L}{2}) x 4 - (\frac{W x L}{2}) x (\frac{L}{4})x 4 \]

\[ Mx4=(({W x Lx2}) x (Lx2)) - ({W x Lx2}) x L \]

Then, if that's right?... could i divide everything by 'L'?

\[ (\frac{Mx4}{L})=(({W x Lx2}) x L) - ({W x Lx2}) \]

I'm not really sure about that, or if it's right, where to go from here. Think I'm getting confused about which parts of the term i should apply the 'action' to (ie. both sides of the '-', or both sides of the 'x' on both sides of the '-')

Sorry, hard to explain in writing but really appreciate your help

Please don't use "x" for times. It can get very confusing. Either just write WL or W*L or [math]W ]cdot L[/math].

[math]M = \left ( \dfrac{WL}{2} ~ \dfrac{L}{2} \right ) - \left ( \dfrac{WL}{2} ~ \dfrac{L}{4} \right )[/math]

[math]M = \dfrac{WL^2}{4} - \dfrac{WL^2}{8}[/math]

Factor the common W:
[math]M = W \left ( \dfrac{L^2}{4} - \dfrac{L^2}{8} \right )[/math]

Can you subtract the fractions?

-Dan

 

[Samwise-zambeezi]

Aaah, think I've got it!

So:

\[ M = W (\frac{L^2}{4}−\frac{L^2}{8}) \]

\[ M = W (\frac{L^2}{8}) \]

\[ M / (\frac{L^2}{8}) = W \]Is that right? Looks right, thanks so much guys!

 

[topsquark]

Otherwise known as [math]W = \dfrac{8M}{L^2}[/math].

Good job!

-Dan