Solve for E

**iamanoobatmath** · September 5th 2008, 02:02 PM

$S=T*\frac{(E-V)}{V}$

My steps....

$VS=T(E-V)$

$TE-TV-VS=0$

$-TV-VS=-TE$

$\frac{(-TV-VS)}{-T}=\frac{-TE}{-T}$

$\frac{(-TV-VS)}{-T}=E$

My question is, is it possible to simplify it further?

**bobak** · September 5th 2008, 02:04 PM

You can factor out the -V giving $E = \frac{V(S+T)}{T}$ .

Bobak

**Matt Westwood** · September 5th 2008, 02:05 PM

First you can multiply top and bottom by -1 to get

$E = \frac{TV+VS}{T}$

then you can extract V as a factor on the top:

$E = \frac{V (T+S)}{T}$

... apart from that, good job.

**iamanoobatmath** · September 5th 2008, 02:09 PM

Thanks for the help

**iamanoobatmath** · September 5th 2008, 03:39 PM

I've gone from
$T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}$
to
$\frac{T^2gh}{4\pi^2}-h^2=k^2$

I'm supposed to solve for k, but I have absolutely no idea how to get rid of the squares.

edit: thanks for the help bobak I figured it out now. I'll just edit this post so I dont' bump the thread
$k=\pm\sqrt{\frac{h(T^2g-4\pi^2h)}{2\pi}}$

**bobak** · September 5th 2008, 04:07 PM

Originally Posted by iamanoobatmath

I've gone from
$T=2\pi\sqrt{\frac{(k^2+h^2)}{gh}}$
to
$\frac{T^2gh}{4\pi^2}-h^2=k^2$

I'm supposed to solve for k, but I have absolutely no idea how to get rid of the squares.

take the square root. so

$k = \pm \sqrt{ \frac{T^2gh}{4\pi^2}-h^2}$

Bobak

**Matt Westwood** · September 5th 2008, 10:24 PM

Originally Posted by iamanoobatmath

$S=T*\frac{(E-V)}{V}$

My steps....

$VS=T(E-V)$

$TE-TV-VS=0$

$-TV-VS=-TE$

$\frac{(-TV-VS)}{-T}=\frac{-TE}{-T}$

$\frac{(-TV-VS)}{-T}=E$

My question is, is it possible to simplify it further?

You can do this easier by:

$VS=T(E-V)$

$VS=TE-TV$ (multiplying out the T)

$TE=VS+VT$ (adding VT (which equals TV) to both sides and changing sides around to put TE on the left)

$TE=V(S+T)$ (factoring by V)

$E = \frac{V(S+T)}{T}$

But it doesn't matter, what you did was perfectly accurate.

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