1. Solving two equations.

I have two equations

$\displaystyle v = (pxy)/m$

$\displaystyle m = (exy)/c^2$

Does that equal?

$\displaystyle v = (pxy)/((exy)/c^2)$

so will those Ys cancel to give:

$\displaystyle v= (pxe)/c^2$

Thanks.

2. Almost, $\displaystyle c^2$ needs to be in the numerator and you need to get rid of $\displaystyle x$ in the numerator since it can cancel along with y. $\displaystyle e$ will remain in the denominator

$\displaystyle v = \dfrac{pc^2}{e}$

out of interest is this to do with physics? $\displaystyle c^2$ tips me off

3. Yes, its to do with momentum Momentum - Wikipedia, the free encyclopedia

Your answer is right but still don't know how to get that. Those are not xs but times. Sorry can't use Latex.

4. Yes, it's to do with momentum Momentum - Wikipedia, the free encyclopedia

Your answer is right but still don't know how to get that. Those are not xs but times. Sorry can't use Latex.

5. Originally Posted by mark090480
$\displaystyle v = (pxy)/m$
$\displaystyle m = (exy)/c^2$
Please do NOT use x as multiplication sign!; use *

p*y / (e*y / c^2)
= p*y * c^2 / (e*y)
= c^2 * p*y / (e*y)
= c^2 * p / e