My Math Forum X(AY+B)+CY+D = (aY+b)(cX+d)

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 November 23rd, 2016, 12:56 AM #1 Newbie   Joined: Nov 2016 From: London Posts: 2 Thanks: 0 X(AY+B)+CY+D = (aY+b)(cX+d) How to calculate a,b,c,d as a function of A,B,C,D, where D is_not_equal BC/A ?? Assume X and Y are variables, and A,B,C,D,a,b,c,d are constants. It can be shown easily that A = ac B = bc C = ad D = bd thus A,B,C,D can be calculated from any a,b,c,d. But what if A,B,C,D are known and D isnot BC/A, how to calculate a,b,c,d ? This question has kept me bogged down for a while and it bothers me that I can't solve it. Your help is appreciated.
 November 23rd, 2016, 04:18 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 2,484 Thanks: 628 If, as you say, B= bc, C= ad, and A= ac then BC/A= abcd/ac= bd= D. It is impossible to find values of A, B, C, D such that D is not BC/A. The problem you are trying to set up is impossible. Thanks from topsquark
 November 23rd, 2016, 08:09 AM #3 Newbie   Joined: Nov 2016 From: London Posts: 2 Thanks: 0 I am afraid you are right. Reason I ask is that I have this formula (left side) and I would like to write it in the form of the right side. But apparently, the equation as per the title is only correct if D=BC/A. So, back to the drawing board for me, and find an alternative for the right side that allows A,B,C,D to be chosen independently. Thanks !
November 27th, 2016, 03:18 AM   #4
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Joined: Nov 2016
From: St. Louis, Missouri

Posts: 28
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Math Focus: arithmetic, fractions
Quote:
 Originally Posted by pin0k10 How to calculate a,b,c,d as a function of A,B,C,D, where D is_not_equal BC/A ?? Assume X and Y are variables, and A,B,C,D,a,b,c,d are constants. It can be shown easily that A = ac B = bc C = ad D = bd thus A,B,C,D can be calculated from any a,b,c,d. But what if A,B,C,D are known and D isnot BC/A, how to calculate a,b,c,d ? This question has kept me bogged down for a while and it bothers me that I can't solve it. Your help is appreciated.
I came up with XAY+XB+CY+D

 Tags bcx, xay

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