# Stumped

• Aug 8th 2007, 10:36 AM
NYKarl
Stumped
Could someone pls show me why:

(m/(n+m))*(V+mK) - mK

is equivalent to:

(m/(n+m))*(V-nK)

Thx for the help
Ben
• Aug 8th 2007, 11:40 AM
ticbol
Quote:

Originally Posted by NYKarl
Could someone pls show me why:

(m/(n+m))*(V+mK) - mK

is equivalent to:

(m/(n+m))*(V-nK)

Thx for the help
Ben

Let's play algebra.

{(m/(n+m))*(V+mK) - mK} =? {(m/(n+m))*(V-nK)}

Notice that [m / (m+n)] is common to both expressions. So we eliminate that to simplify the two expressions. Divide both sides by [m / (m+n)],

(V +mK) -(mK / [m /(m+n)]) =? V -nK

Then simplify. Expand the Lefthand Side,

(V +mK) -(mK * [(m+n) / m]) =? V -nK
(V +mK) -[(mK)(m+n) / m] =? V -nK
Clear the fraction, multiply both sides by m,
m(V +mK) -mK(m+n) =? mV -mnK
mV +(m^2)K -(m^2)K -mnK =? mV -mnK
mV -mnK =? mV -mnK
Yes.

So, it is shown.
• Aug 8th 2007, 01:01 PM
CaptainBlack
Quote:

Originally Posted by NYKarl
Could someone pls show me why:

(m/(n+m))*(V+mK) - mK

is equivalent to:

(m/(n+m))*(V-nK)

Thx for the help
Ben

$\displaystyle \frac{m}{n+m}(V+mK) - mK = \frac{mV+m^2K-nmK-m^2K}{n+m}$$\displaystyle =\frac{mV-nmK}{n+m}=\frac{m}{n+m}(V+nK)$

RonL