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I am looking at a question involving three equations:

  1. $A=0.6667A+0.2222B+0.1667C$
  2. $B= 0.2A+0.3333B+0.5C$
  3. $C=0.1333A+0.4444P+0.3333C$

The solution then goes on to say, that these equations can be re-written as:

  1. $0.33A+0.22B+0.17C=0$
  2. $-0.2A+0.67B+0.5C=0$
  3. $-0.13A-0.44B+0.67C=0$

Can anyone explain how you get the second set of equations from the first? Pulling my hair out!!! If you take $A,B$ & $C$ from LHS over to RHS, then the signs change etc, but I don't get the second set at all.

One other equation that goes along with these are that $A+B+C=1$

Any help would be appreciated

Em.
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Rab1
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1 Answers1

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Serval things happened. I will only consider equation 1.

$$A=0.6667A+0.2222B+0.1667C$$

First I subtract $0.6667A$ from both sides so that on the LHS I get $$A-0.6667A = (1-0.6667)A = 0.3333A$$ which was rounded to $0.33A$. Moving the other two terms to the LHS and more rounding gives $$0.33A+0.22B+0.17C=0$$

which is equation 4. Similar rounding occurs in the other equations.

Em.
  • 15,981
  • Hi thanks for the reply...I get the first part, but how do you end up with +0.22A + 0.17C......Can you explain the steps you took please – Rab1 Jul 11 '16 at 13:20