small bowl           c
large bowl           12
lid                  x

A 2 bowl set has the following description:

The lid is interchangeable.

Weight of larger bowl:  12 oz. without lid.

With lid, larger weighs twice as much as smaller without lid.

With lid, smaller weighs one third more than larger without lid.

Calculate weight of lid.

Let A = larger.

Let B = smaller.

Let C = lid.

A = 12.

A + C = 2(B).

A = 1/3(B + C)

Not sure if set up correctly and how to solve.

Why did that other person bring in another variable, x? Doesn't the problem list enough variables?

A = 1/3(B + C) <==== This is WRONG!! Read on!

Weight of larger bowl, without lid: A = 12
Weight of larger bowl, with lid: A + C, or 12 + C

Weight of smaller bowl, without lid: B
Weight of smaller bowl, with lid: B + C

As larger, with lid, weighs twice as much as smaller without lid, we get:  --- eq (i)

    As smaller, with lid, weighs  MORE than larger, without lid, we get: 
                                                                            -- Substituting 12 for A                                          
                                                                           B + C = 16
                                                                               B = 16 - C ----- eq (ii)

                                                                   2(16 - C) - C = 12 ---- Substituting 16 - C for B in eq (i)
                                                                     32 - 2C - C = 12
                                                                        - 2C - C = 12 - 32                                                                        
                                                                            - 3C = - 20
                                                            Weight of lid, or