Solve each of the following systems by addition.
2x+3y =  1
5x+3y = 16

The idea is to multiply each equation by whatever number
will be such that when you add the equations vertically,
one of the letters will cancel out.  The 3y in the top can be
made to cancel with the 3y in the bottom if one of them
became -3y and the other remained +3y.  So we multiply the
first equation through by -1 and the second equation through
by +1:


-1[2x+3y =  1]
+1[5x+3y = 16]

This becomes:

  -2x-3y = -1
   5x+3y = 16

Now we draw a line underneath and add
vertically term by term:

  -2x-3y = -1
   5x+3y = 16
   ----------
   3x    = 15

Which we solve and get

       x = 5

Now you can finish by switching over to
substition, which is what many teachers
and books tell you to do.  However, to 
finish by addition, start over and 
eliminate the x terms:

2x+3y =  1
5x+3y = 16

The 2x in the top can be made to cancel with the 5x in the 
bottom if one of them becomes -10x and the other becomes
+10x, since 10 is the LCM of 2 and 5.  So we multiply the 
first equation through by -5 and the second equation through 
by +2:

-5[2x+3y =  1]
+2[5x+3y = 16]

This becomes:

 -10x-15y = -5
  10x+ 6y = 32

Now we draw a line underneath and add
vertically term by term:

 -10x-15y = -5
  10x+ 6y = 32
   -----------
      -9y = 27

Which we solve and get

        y = -3

So the solution is 

(x,y) = (5,-3)

Edwin