SOLUTION: Solve the linear system...if the system is inconsistent or dependent, so state: 3x+2y+z=2 4x-y+3z=-16 x+3y-z=12

3x + 2y +  z =   2
4x -  y + 3z = -16
 x + 3y -  z =  12

1. If there is an equation with only two unknowns in it, then use it, and
skip to step 3.  Otherwise, create an equation in only two unknowns by using
step 2.

There is no equation in only two unknowns, so we must go to step 2
  
2. Arbitrarily pick two equations, and an unknown to eliminate from them.
Although it really doesn't matter which two equations you pick, you will
save time by picking two equations which have an unknown easy to eliminate.

I will arbitrarily pick the first and third, and unknown z to eliminate,
because I can just add them as they are and the z's will cancel  

3x + 2y + z =  2
 x + 3y - z = 12
————————————————
4x + 5y     = 14

3. Pick a pair of equations (not the same pair you picked in step 2 (if you did
step 2), and eliminate from them the SAME unknown which is missing in your
equation in only two unknowns you got from either step 1 or 2.

I will pick the 2nd and 3rd equations, and I have no choice but to eliminate
z from them, bacause that is the unknown that is missing in the equation I 
got from step 2. 

4x -  y + 3z = -16
 x + 3y -  z =  12

To eliminate z, multiply the second equation by 3 and add to the first
equation:

4x -  y + 3z = -16
3x + 9y - 3z =  36
——————————————————
7x + 8y      =  20

4. Solve the pair of equations in two unknowns for the unknowns.
You may use either substitution of addition-subtraction/elimination
method:

Solve the system by whatever method you like best. If you can't solve
a system of two equations in two unknowns, then post again.  Assuming
you know how to solve this system:

4x + 5y = 14
7x + 8y = 20  

I'll just give you this answer:  you get x = -4 and y = 6

5.  From the original system, arbitrarily pick an equation that contains the
other unknown you still don't have the value of.  Substitute the values for
the two unknowns you now have and solve for the final letter.

I arbitrarily pick the first original equation

3x + 2y + z = 2 and replace x by -4 and y by 6

3(-4) + 2(6) + z = 2

-12 + 12 + z = 2
           
           z = 2

So the final solution is (x, y, z) = (-4, 6, 2)

Edwin McCravy
AnlytcPhil@aol.com