Question 249564
let x = number of singers in each row.
let y = number of rows.


at BC we have x*y = 330


at Coquitlam we have (x-4)*(y+1) = 319


we need to solve both these equations simultaneously to get our answer.


use the first equation to solve for one of the variable in terms of the other.


you get y = 330/x


substitute for y in the second equation to get:


(x-4) * (330/x + 1) = 319


simplify by performing indicated operations to get:


330 + x - (4*330)/x - 4 = 319


multiply both sides of the equation by x to get:


330*x + x^2 - (4*330) - 4*x = 319*x


subtract 319*x from both sides of the equation to get:


330*x + x^2 - (4*330) - 4*x - 319*x = 0


simplify by performing indicated operations to get:


330*x + x^2 - 1320 - 4*x - 319*x = 0


simplify by combine like terms to get:


x^2 + 7*x - 1320 = 0


factor to get:


(x+40) * (x-33) = 0


solve for x to get:


x = -40 and x = 33


since x can't be negative, x must equal 33.


x = 33 means there were 33 singers in each row at BC.


this means y = 10 which means there were 10 rows at BC.


at Coquitlam they had 4 less singers in each row and 1 more row.


This makes 29 singers and 11 rows at Coquitlam.


29 * 11 = 319 singers total at Coquitlam.


Everything checks out so your answer is:


there were 10 rows of singers at BC.