SOLUTION: Solve each system by elimination. 10) x+y+z=-2 2x+2y-3z=11 3x-y+z=4

 x +  y + z =  -2    (1)
2x + 2y - 3z = 11    (2)
3x -  y +  z =  4    (3)
~~~~~~~~~~~~~~~~~~~~~~~~~~~



You are exceptionally lucky person, because in this case the solution 
of the given system is very easy.


Multiply equation (1) by 2 (both sides); then subtract from equation (2).

You will get then  

    -3z - 2z = 11 - 2*(-2)

    - 5z = 15

       z = 15/(-5) = -3.


   +---------------------------------------+
   |  First elimination step is complete.  |
   +---------------------------------------+


Now substitute z= -3 into equations (1) and (3).  You will get then

    x +  y + (-3) = -2    (4)

    3x - y + (-3) =  4    (5)


Add equations (4) and (5).  You will get

    4x + (-6) = 2  --->  4x - 6 = 2  --->  4x = 2 + 6 = 8  --->  x = 8/4 = 2.


   +----------------------------------------+
   |  Second elimination step is complete.  |
   +----------------------------------------+


Now substitute x= 2 and z= -3 into equation (1).  You will get

     
    2 + y + (-3) = -2  --->  y = -2 - 2 + 3 = -1.


ANSWER.  The solution is  x= 2,  y= -1,  z= -3.


To check, substitute these values to each equation  and make sure that left sides are equal to right sides.