Question 90432



      Solve the system by addition or substitution.
3x – 4y = 8
6x – 2y = 10

       3x-4y = 8.........(1)

       6x- 2y = 10.......(2) 


       From eq'n (2)  y = (10-6x)/-2 = -5+3x  

      substituing the value of  y in eq'n (1)  we get 3x-4(-5+3x) = 8

                    3x+20-12x = 8
                    -9x = 8-20 = -12
                      x = -12/-9 = 4/3
          substitute the value of x in eq'n (1)   we get 3(4/3)-4y = 8

                                                   4-4y = 8
                                                   -4y = 8-4 = 4
                                                     y = 4/-4 = -1

            The values of x = 4/3 and y = -1  are the solutions for the eq'ns