Posted: 26 October 2007 at 10:04am  IP Logged



First to solve this problem by using elimination you are going to need to choose which variable (letter) you want to work with. I choose "z" because the only thing you will need to do is mulitple the equation(s) by a negative instead of a negative and a number like "x" and "y".
So, let's make pairs to eliminate z: I choose (1,2) (1,3)
2xy+z=3 (1 stays the same but what do we have to do to 2 to eliminate "z".. multiply it by a )
(x+2y+z=12)
So the new equation becomes
2xy+z=3
x2yz=12
Now add them together and ''z'' is gone (x3y=9)
Now, do the same for (1,3) and you should get (2x+2y=2)
Now take x3y=9 and 2x+2y=2 and solve for "x" or "y".
"x" should = 3 and "y" should = 4
Take and plug both of those in the original equation (1, 2, or 3) and solve for "z".
You should get "z" = 1
Now, check this on all the equations by plugging in the values.
Example 2(3)(4)+1=3 AKA 2xy+x=3
Answer
(3,4,1)
