Solving two equations.

**mark090480** · March 8th 2011, 05:31 AM

I have two equations

$v = (pxy)/m$

$m = (exy)/c^2$

Does that equal?

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

so will those Ys cancel to give:

$v= (pxe)/c^2$

Thanks.

**e^(i*pi)** · March 8th 2011, 05:37 AM

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

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

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

**mark090480** · March 8th 2011, 05:48 AM

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.

**mark090480** · March 8th 2011, 05:49 AM

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.

**Wilmer** · March 8th 2011, 09:10 AM

Originally Posted by mark090480

$v = (pxy)/m$
$m = (exy)/c^2$

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

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