My teacher today solved this system of equations for us that consisted of these 3 equations,

1)  p0   +   p1   =1

2) a*p0  +  b*p1  =p0

3) c*p0  +  d*p1  =p1

, like this!

First he wrote this equation:

p0= (p1 - d*p1)/c

Second he wrote this equation(THE PROBLEM):

c= p1 - d*p1 + c*p1

And last was this equation :

p1= c/(1 - d + c) 

He did it so fast without really explaining anything and only after lesson I understood that I do not know how did he got the second equation!

Can anyone please explain how did he got the 2nd equation in his offered solution????

  • $\begingroup$ multiply the first one by $c$ and replace $cp_0$ thanks to the third one $\endgroup$ – Vincent May 5 '16 at 21:19
  • $\begingroup$ @Vincent if doing so I do not get cp1! How can i get cp1? $\endgroup$ – Mārcis Liepiņš May 5 '16 at 21:24
  • $\begingroup$ The first equation is $p_0 + p_1 = 1$ solve this for $p_0$ and plug it into the first equation your teacher derived. Then solve for c. $\endgroup$ – jazzinsilhouette May 5 '16 at 21:26
  • $\begingroup$ I am talking about equation 1), not the first on your list that comes later $\endgroup$ – Vincent May 5 '16 at 21:27
  • $\begingroup$ @Vincent Thanks! Worked! $\endgroup$ – Mārcis Liepiņš May 5 '16 at 21:36

Rewrite equation 1) by isolating p0. Call the new equation 1a).

p0 + p1 = 1   ==>  p0 = 1 - p1

And rewrite 3) by isolating cp0*. Call the new equation 3a).

c*p0 + d*p1 = p1   ==>  c*p0 = p1 - d*p1

Now substitute the value for p0 found in 1a) into equation 3a).

c*p0 = y - d*p1  ==> c*(1 - p1) = y - d*p1

Simply distribute and isolate c and you are done.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.