Originally Posted by

**TreeMoney** I set it up so i had 3 equations they were

3 = x1 + 3x2 +2x3

-1 = -x1 + x2 + 2x3

2 = 2x1 - 2x2 - 4x3

Replace the second equation by the second plus the first, give us:

Code:

3 = x1 + 3.x2 +2.x3
2 = + 4.x2 +4.x3
2 = 2.x1 - 2.x2 - 4.x3

Now replace the third row by the third row minus twice the first

row:

Code:

3 = x1 + 3,x2 +2.x3
2 = + 4.x2 +4.x3
-4 = - 8.x2 - 8.x3

Now we see that the last two rows are multiples of one another,

which leaves us with two effective equations in three unknowns.

This leaves us with multiple solutions so assume a value for x3

and then solve for x1 and x2 in terms of this.

For instance x1=3/2, x2=1/2, x3=0 looks like a solution to me

(assuming no arithmetic errors).

RonL