Question 88727
   1)x/3+4y = 30......eq'n(1)
     2x-y = 5 ........eq'n(2)
   multiply  eq'n(1) by 3 throughout  we get x+12y = 90 hence x = 90-12y
                 substitute for x in eq'n (2)  we get 2(90-12y)-y = 5
                                                       180- 24y-y = 5
                                                       180-5  = 25y
                                              25y = 175  or y = 175/25
                                                            y = 7
                 substitute for y in 2x-y = 5
                                      2x = 5+y = 5+7 = 12 i.e,x = 12/2 = 6
        the solution for the eq'ns are x = 6   and y = 7

   2)6x-y/4 = -9........eq'n (1)
     3x+2y = -30........eq'n(2)

     multiply   eq'n (1) by 4 throughout  we get 24x-y = -36 
                                                 24x+36 = y
     substitute for y in eq'n(2)    we get 3x+2(24x+36) = -30
                                           3x+48x+72= -30
                                                  51x = -30-72 = -102
                                                    x = -102/51 = -2
      substituting for x in 3x+2y =-30   we get 3(-2)+2y = -30
                                                -6 +2y -30 or 2y = -30+6 = -24
                                                              y = -24/2 = -12 
      the solution for the eq'ns is x = -2  and y = -12

    3)y = x^2+2x-2.........eq'n (1)
      y = 3x+4 ............eq'n(2)
          substituting the value of y from eq'n(2)  in eq'n(1)
      we get 3x+4 = x^2+2x-2  or  x^2+2x-3x-2-4 = 0
                                  x^2-x-6 = 0    spilliting -6 = -3*+2
                                  x^2-3x+2x-6=0
                                  x(x-3)+2(x-3) = 0
                                      (x-3).(x+2) = 0 or x-3 = 0, x=3
                                                         x+2 = 0 , x = -2 

          if x=-2then value of y = 3(-2)+4 = -6+4 = -2
          if x = 3 then the value of y = 3(3)+4 = 9+4 = 13