Originally Posted by
CaptainBlack Multiply the first equation by 2 and subtarct the second to get:
(2x+4y+2az)-(2x+3y-2z)=0-a
or:
y + (2a+2)z=-a. ...(A)
Now multiply the second equation by a and the third by 2 and subtarct:
(2ax + 3ay - 2az)-(2ax + 2y + 2z) = a^2-6
or:
(3a-2)y - (2a+2)z = a^2-6 ...(B)
Now add equationa (A) and (B) to get:
(3a-1)y = a^2-a-6,
so:
y = (a^2-a-6)/(3a-1) ...(C).
Substituting this into equation (A) gives:
z=-(4a^2-2a-6)/[(3a-1)(2a+2)]=-[(a+1)(2a-3)]/[(3a-1)(a+1)]
...........................=-(2a-3)/(3a-1) ...(D)
Then substitute (C) and (D) into the first equation to find x.
RonL
(Check the algebra it can always go wrong)