# Solving two equations.

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• Mar 8th 2011, 05:31 AM
mark090480
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.
• Mar 8th 2011, 05:37 AM
e^(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 AM
mark090480
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 AM
mark090480
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 AM
Wilmer
Quote:

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