The original arrangement of the class is S students in each of R rows. So SR = the number of students in the class. The first new arrangement of the class is S+1 students in each of R-2 rows. So (S+1)(R-2) also equals the number of students in the class. The second new arrangement of the class is S-1 students in each of R+3 rows. So (S-1)(R+3) also equals the students in the class. Since the number of students is the same under each of the three arrangements, SR = (S+1)(R-2) = (S-1)(R+3) SR = SR-2S+R-2 = SR+3S-R-3 Subtract SR from all three sides: 0 = -2S+R-2 = 3S-R-3 So each of those are equal to 0 -2S+R-2 = 0 3S-R-3 = 0 Adding the two equations term by term: S -5 = 0 S = 5 Substituting S = 5 in -2S+R-2 = 0 -2(5)+R-2 = 0 -10+R-2 = 0 -12+R = 0 R = 12 So the number of students is SR = (5)(12) = 60 Checking. In the original arrangement there were 5 students in each of 12 rows. In the first new arrangement there were 6 students in each of 10 rows. That's 1 student more and 2 rows less than in the original arrangement. In the second new arrangement there were 4 students in each of 15 rows. That's 1 student less and 3 rows more than in the original arrangement. Edwin