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Question
Let P = a2 − b2 + 2ab, Q = a2 + 4b2 − 6ab, R = b2 + 6, S = a2 − 4ab and T = −2a2 + b2 − ab + a. Find P + Q + R + S − T.
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Solution
P = a2 − b2 + 2ab
Q = a2 + 4b2 − 6ab
R = b2 + 6
S = a2 − 4ab
T = −2a2 + b2 − ab + a
Adding P, Q, R and S:
P+Q+R+S
= (a2 − b2 + 2ab)+(a2 + 4b2 − 6ab)+(b2 + 6)+(a2 − 4ab )
= a2 − b2 + 2ab+a2 + 4b2 − 6ab+b2 + 6+a2 − 4ab
Rearranging and collecting the like terms:
= (1+1+1)a2 +(−1+4+1) b2 + (2-6-4)ab+6
P+Q+R+S = 3a2 +4b2 - 8ab+6
To find P + Q + R + S − T, subtract T = (−2a2 + b2 − ab + a) from P+Q+R+S = (3a2 +4b2 - 8ab+6).
On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a2 + b2 − ab + a
Changing the sign of each term of the expression gives 2a2 - b2 + ab - a.
Now add:
(3a2 +4b2 - 8ab+6)+(2a2 - b2 + ab - a) = 3a2 +4b2 - 8ab+6+2a2 - b2 + ab - a
= (3+2)a2 +(4-1) b2 +(-8+1) ab - a+6
P + Q + R + S − T = 5a2 +3b2 -7 ab - a+6
Q = a2 + 4b2 − 6ab
R = b2 + 6
S = a2 − 4ab
T = −2a2 + b2 − ab + a
Adding P, Q, R and S:
P+Q+R+S
= (a2 − b2 + 2ab)+(a2 + 4b2 − 6ab)+(b2 + 6)+(a2 − 4ab )
= a2 − b2 + 2ab+a2 + 4b2 − 6ab+b2 + 6+a2 − 4ab
Rearranging and collecting the like terms:
= (1+1+1)a2 +(−1+4+1) b2 + (2-6-4)ab+6
P+Q+R+S = 3a2 +4b2 - 8ab+6
To find P + Q + R + S − T, subtract T = (−2a2 + b2 − ab + a) from P+Q+R+S = (3a2 +4b2 - 8ab+6).
On changing the sign of each term of the expression that is to be subtracted and then adding:
Term to be subtracted = −2a2 + b2 − ab + a
Changing the sign of each term of the expression gives 2a2 - b2 + ab - a.
Now add:
(3a2 +4b2 - 8ab+6)+(2a2 - b2 + ab - a) = 3a2 +4b2 - 8ab+6+2a2 - b2 + ab - a
= (3+2)a2 +(4-1) b2 +(-8+1) ab - a+6
P + Q + R + S − T = 5a2 +3b2 -7 ab - a+6
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