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.

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- Mar 8th 2011, 05:31 AMmark090480Solving 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. - Mar 8th 2011, 05:37 AMe^(i*pi)
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 - Mar 8th 2011, 05:48 AMmark090480
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. - Mar 8th 2011, 05:49 AMmark090480
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. - Mar 8th 2011, 09:10 AMWilmer