Question

3a+ 6 = 15, 3a = 9, a+ 2= 5, 1/3a =1

Answer 1

Given parameters:

  Find if any of the equations are the equivalent;

           3a+ 6 = 15

            3a = 9

            a+ 2= 5

              $\frac{1}{3a}$ =1

To find which of the equations are equivalent to one another, let us solve them first. The ones with the same solution are definitely equivalent.

     (i)  3a+ 6 = 15

                3a  = 15 - 6

                 3a  = 9

                    a = 3

       (II)   3a = 9

                a = 3

    (III)     a+ 2= 5

                a = 5-2

                a = 3

   (IV)      $\frac{1}{3a}$ =1

             1 = 3a

             a  = $\frac{1}{3}$

We clearly see that the first three equations are equivalent.