SOLUTION: I am supposed to solve the systems of equation in three variables. 5x+2y=4 3x+4y+2z=6 7x+3y+4z=29

Equation (1):     5x+2y = 4
Equation (2):     3x+4y+2z = 6
Equation (3):     7x+3y+4z = 29

Since z is already eliminated in the equation (1), we
eliminate z from equations (2) and (3) by multiplying
equation (2) through by -2 getting

                 -6x-8y-4z = -12

and adding it to equation (3):

                  7x+3y+4z =  29
                 -6x-8y-4z = -12
                 ---------------
Equation (4)       x-5y    =  17

Now we take equation (1) with equation (4):

Equation (1):     5x+2y =  4
Equation (4)       x-5y = 17

Solve equation (4) for x:

Equation (5)          x = 17+5y

and substitute 17+5y for x in equation (1)

            5(17+5y)+2y = 4
              85+25y+2y = 4
                 85+27y = 4
                    27y = 4-85
                    27y = -81
                      y = -3

Substituting that in equation (5):

                      x = 17+5y
                      x = 17+5(-3)
                      x = 17-15
                      x = 2

Substitute x = 2 and y = -3 in equation (2)

               3x+4y+2z = 6
          3(2)+4(-3)+2z = 6
                6-12+2z = 6
                  -6+2z = 6
                     2z = 12
                      z = 6

So the solution is , , 

Edwin