Looking for an identity or process to break one function into to

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CD 2019-2-8

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评论： Walter Roberson 2019-2-8

Would like to go from here:

(A'*C - B*C) / (A + B) to

A'*C/A - B*C/(A*(A+B))

Can anyone name this identity or process which pulls A'*C/A out of (A'*C - B*C) / (A + B) as shown above?

Thank you.

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CD 2019-2-8

I should have been more clear. I forgot to point out that there is a value "A" and value "A' " where A' = 1 -A.

I've tested this in excel:

A A' B C

0.2 0.8 2 3

(A'C - BC) / (A + B)

-1.636

A'C/A - BC/(A(A+B)

-1.636

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Answer 1

Walter Roberson 2019-2-8

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在 MATLAB Online 中打开

False. Suppose A = 5 and B = 11 then

>> syms C
>> (5*C - 11*C)/(5+11)
 
ans =
 
-(3*C)/8
 
>> 5*C/5 - 11*C/(5*(5+11))
 
ans =
 
(69*C)/80

Not even the same sign.

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Walter Roberson 2019-2-8

在 MATLAB Online 中打开

Consider

(Ap*C - B*C)/(A + B) = Ap*C/A - B*C/(A*(A + B))

multiply both sides by A/C to get

(Ap*C - B*C)*A/((A + B)*C) = (Ap*C/A - B*C/(A*(A + B)))*A/C

On both sides, the C cancel in the top and bottom. On the right side th A cancel on the top and bottom

(Ap - B)*A/(A + B) = Ap - B/(A + B)

normalize the right side into a fraction

(Ap - B)*A/(A + B) = (A*Ap + Ap*B - B)/(A + B)

discard the denominator

(Ap - B) * A = A*Ap + Ap*B - B

expand

Ap*A - A*B = A*Ap + Ap*B - B

cancel Ap*A on both sides:

-A*B = Ap*B - B

factor right side:

-A*B = (Ap-1)*B

Ap is 1-A so (Ap-1)*B is (1-A-1)*B = -A*B which is the left hand side. Therefore the two sides are equal -- except for the cases where A=-B or A = 0 or C = 0, which would have to be examined more carefully to avoid multiplication or division by 0 giving false equations.

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