# Thread: Solving a set of linear equations problem

1. 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)
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

2. Originally Posted by feage7
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
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

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