We can do the problem with or without algebra.  We'll do it both ways.

Without algebra:
Tables for 4 are tables for 2 couples and tables for 6 are tables for 3 couples.
There are seats for 152 patrons, or for 152÷2 or 76 couples. If all 32 tables
were tables for 2 couples there would only be seats for 32×2 or 64 couples.
So the other seats for the 76-64 or 12 couples must be at tables for 3 couples.
So there are 12 tables for 6 (and 32-12 or 20 tables for 4).

Answer: 12 tables for six.

With algebra.

Make this chart:

            number       seats      number or
           of tables   per table      seats      
---------------------------------------------
for four    32-x          4          4(32-x)
for six       x           6            6x
---------------------------------------------
totals        32                      152

The equation comes from:




    

Solve that and get x=12, or 12 tables for six, (and 32-12=20 tables for four). 

Edwin