# Have do i get 2nd equation from 3 equation 2 variable system answer?

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


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????

• multiply the first one by $c$ and replace $cp_0$ thanks to the third one – Vincent May 5 '16 at 21:19
• @Vincent if doing so I do not get cp1! How can i get cp1? – Mārcis Liepiņš May 5 '16 at 21:24
• 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. – jazzinsilhouette May 5 '16 at 21:26
• I am talking about equation 1), not the first on your list that comes later – Vincent May 5 '16 at 21:27
• @Vincent Thanks! Worked! – 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.