When I followed this through i didnt get the /(3a-1) bit in equation D was hoping you could break it down a bit more so I can see where it came from

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- Nov 10th 2006, 12:38 AMfeage7
- Nov 10th 2006, 04:08 AMCaptainBlack
We have:

y + (2a+2)z=-a. ...(A)

y = (a^2-a-6)/(3a-1) ...(C)

substituting (C) into (A) gives:

(a^2-a-6)/(3a-1) + (2a+2)z=-a,

so:

(2a+2)z=-a - (a^2-a-6)/(3a-1) = -[a(3a-1)+(a^2-a-6)]/(3a-1)

...........= -[4a^2-2a-6]/(3a-1)=[(a+1)(4a-6)]/(3a-1)

so:

z = (2a-3)/(3a-1)

RonL