<PRE><B>Question 98855</B>


   solve this system of equations with elimination:
A: x-2y+3z=4
B: 2x+y-4z=3
C: -3x+4y-z=-2 

   To solve the given eq&#39;ns by elimination we have to first solve A and B  then solve Band C  and then solve the resulting eq&#39;ns

     A: x-2y+3z = 4
     B: 2x+y-4z = 3  multiply B by 2  B: 4x+2y-8z = 6
                                      A:  x-2y+3z = 4
                                        ---------------
                                         5x   -5z = 10
                                        --------------
          5x-5z = 10   or 5(x-z) = 10   or x-z = 10/5 = 2........(D)

     B:2x+y-4z = 3
     C:-3x+4y-z = -2   multiply B  by 4  B:8x+4y-16z = 12
                                        C: -3x+4y-z  = -2

                     subtracting          -11x-15z = 14......(E)

          from D substitute for x in (E)  we get -11(2+z)-15z = 14
                                                -22-11z-15z = 14
                                                -26z = 14+22 =36
                                                   z = -36/26
                              from (D)  x-z = 2
                                        x = 2+z = 2-36/26
                                                = 52-36/26 = 16/26
                                             x = 16/26

         substituting for x and z in B  we get 2(16/26)+y-4(-36/26)= 3
                                              32/26 +y +144/26 = 3
                                                    y + 176/26 = 3
                                                       y = 3-176/26
                                                         = 78-176/26
                                                         = -98/26

     the values of x,y and z is x = 8/13, y = -49/13 and z = -18/13



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