中二恆等式 唔該幫幫忙

P(a-2)-(a-Q)=6a-7
Pa+(Q-2P)=7a-7

Pa=7a
P=7

Q-2P=-7
Q-2*7=-7
Q=7

答案是：
P(a-2)-(a-Q)=6a-7
Pa+(Q-2P)=7a-7

Pa=7a
P=7

Q-2P=-7
Q-2*7=-7
Q=7

P(a-2)-(a-Q)=6a-7
L.H.S. = P(a-2)-(a-Q)
=Pa- 2P -a + Q
=(P-1)a + (-2P + Q)
=(P-1)a - (2P - Q)
∴(P-1)a - (2P - Q) = 6a-7
By comparing the like terms on L.H.S. and R.H.S.,
Therefore, P-1 = 6
P = 7
Hence, 2P - Q = 7
2(7) - Q = 7
Q = 14 - 7
Q = 7

P(a-2)-(a-Q)=6a-7
aP-2P-a+Q=6a-7
aP-a-2P+Q=6a-7
(aP-a)+(Q-2P)=(6)a+(-7)
(P-1)a-(Q-2P)=(6)a+(-7)



∴ P-1=6
P=5
Q-2P=-7
Q-2(5)=-7
Q-10=-7
Q=3

2009-11-02 16:42:12 補充：
P-1=6
P=7

Q-2P=-7
Q-2(7)=-7
Q-14=-7
Q=7