Question 12654
Using your third equation, solve for X and substitute it in for X in your other two equations...<BR>
X = 3y - 4z - 14<BR>
3(3y - 4z - 14) + 2y - z = 8     
   9y - 12z - 42 + 2y - z = 8       
   11y - 13z - 42 = 8            
   11y - 13z = 50 
<BR>
-3(3y - 4z - 14) + 4y + 5z = -14 
     9y - 12z - 42 + 2y - z = 8  
    -9y + 12z + 42 + 4y + 5z = -14 
    -5y + 17z + 42 = -14 
<BR>
Take these two equations and use linear combinations<BR>

   5(11y - 13z = 50)                    
  11(-5y + 17z = -56)
<BR>
   55y -  65z = 250
+ -55y + 187x = -616
----------------------
         122z = -366  ----> <B> ( Z = -3 ) </B>
<BR>
Now that we know Z = -3, substitute it back into two of your original equations.
<BR>
3x + 2y - (-3) = 8              
3x + 2y + 3 = 8                                     
-3x + 4y = 1
<BR>
-3x + 4y + 5(-3) = -14
-3x + 4y - 15 = -143x + 2y = 5 
<BR>
Use these to do Linear Combinations
    3x + 2y = 5
+  -3x + 4y = 1
------------------
         6y = 6     ------> <B> ( Y = 1 ) </B>
<BR>
Now Take your Original Equation 3x+2y-z=8 Substitute Z and Y to find X
   3x + 2(1) - (-3) = 8
   3x + 2 + 3 = 8
   3x + 5 = 8
   3x = 3
    x = 1

<P>
So Z = -3, Y = 1, and X = 1