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Math is not my strong suit. As far as I can tell, this is what I'm looking to solve.

A+B=C and B=C*D

If one knows what A and D equal, can one determine the value of B and C?

So far, I've tried:

A+B=C
A+C*D=C
C*D=C-A
C*D-C=-A
A=C-C*D

And then I get stuck. Also:

A+B=C
B=C-A
C*D=C-A
C=(C-A)/D
1=((C-A)/D)/C

Stuck again.

Is this doable?

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2  
So close.... You got to $CD-C=A$; can you do something to the left side? –  Gerry Myerson Jul 20 '13 at 6:29
    
It's killing me - I can't figure out how to get C alone on the left side! –  Eric Oliver Jul 20 '13 at 6:35
1  
Sorry, I meant $CD-C=-A$; now do something just to the left side. –  Gerry Myerson Jul 20 '13 at 6:35
2  
So if CD-C=-A then C(D-1)=-A then C=-A/(D-1) Is that right? –  Eric Oliver Jul 20 '13 at 6:50
1  
Eric, once you have figured the whole thing out, you can post it as answer, and then, later, accept it. That helps cut down the Unanswered Questions queue. –  Gerry Myerson Jul 20 '13 at 6:57
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1 Answer 1

up vote 0 down vote accepted

From eqn. 1 put the value of $C$ in eqn. 2 and you get \begin{equation} \begin{split} B=&(A+B)D \\ \Rightarrow \ B-BD=&AD\\ \Rightarrow B=\frac{AD}{1-D} \end{split} \end{equation}. The last step follows by taking the $BD$ in RHS to LHS and then taking some common factor. Since you know $A,\ D$, so you know B. Now put this value of $B$ in eqn. 1 to get $$C=A+\frac{AD}{1-D}=\frac{A}{1-D}$$

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3  
Why not let OP work this out? He's so close.... –  Gerry Myerson Jul 20 '13 at 6:56
    
Ok, yeah true @GerryMyerson –  Samrat Mukhopadhyay Jul 20 '13 at 6:57
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