SOLUTION: Hello! I'm making my way through a book called "The Mathematics of Games and Gambling" by Edward Packel. There's an algebraic equation therein which goes like this: x

Question 28775: Hello! I'm making my way through a book called "The Mathematics of Games and Gambling" by Edward Packel. There's an algebraic equation therein which goes like this:
x=4pq-5(1-q)-2(1-p)q+3(1-p)(1-q). The author partially reduces this expression to the following:
x=14pq-8p-5q+3. I don't understand how he arrived at that point. Using my own highly dodgy math skills, I came up with the following:
x=9pq-5p+2q-2. I tried the "collecting like terms" operation, but I obviously veered off the right path somewhere. Could someone show me the light here, please? With many thanks, RR
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
OH! MY DEAR !! I THINK YOU MADE A SIMPLE MISTAKE IN COPYING PROBABLY..SEE BELOW
Hello! I'm making my way through a book called "The Mathematics of Games and Gambling" by Edward Packel. There's an algebraic equation therein which goes like this:
x=4pq-5(1-q)-2(1-p)q+3(1-p)(1-q).
I THINK YOU MISSED P IN THE SECOND TERM .I PUT IN CAPITAL LETTER IN BRACKETS.IT SHOULD BE
x=4pq-5(P)(1-q)-2(1-p)q+3(1-p)(1-q)...SO NOW WE GET
X=4PQ-5P+5PQ-2Q+2PQ+3-3P-3Q+3PQ
=14PQ-8P-5Q+3...OK.....
The author partially reduces this expression to the following:
x=14pq-8p-5q+3.
OK......SAME AS ABOVE
I don't understand how he arrived at that point. Using my own highly dodgy math skills, I came up with the following:
x=9pq-5p+2q-2.
YOU MADE A MISTAKE HERE..JUST SEE
x=4pq-5(1-q)-2(1-p)q+3(1-p)(1-q).
=4PQ-5+5Q-2Q+2PQ+3-3P-3Q+3PQ
=9PQ-3P-2....HOW DID YOU GET x=9pq-5p+2q-2. ?..CHECK....

I tried the "collecting like terms" operation, but I obviously veered off the right path somewhere. Could someone show me the light here, please? With many thanks, RR

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