A city bus collected $1,140 in fares on a given day. The regular fare is $1.50, and the reduced fare is $0.75. If 1,042 total fares were paid, which system of linear equations can be used to find x, the number of regular fares paid, and y, the number of reduced fares paid?

x + y = 1042

1.50x + 0.75y = 1140

Step-by-step explanation:

If there was a total of 1,042 fares paid then x + y would be equal to that because x is the number of regular fares paid and y is the number of reduced fares paid.

If there was a total of $1,140 made from the fares then we can use 1.50x + 0.75y. We put the 1.5 in front of the x because each fare paid costs $1.50, same with Y each reduced fare paid is $0.75. So both of those need to be added up to get $1,140.

(n^3 - n)

= n(n-1)(n+1)

since it is divisible by 3 and 2 ,because it is a multiple 3 consecutive number

if, n-1 is odd then n+1 is also odd, then it is a multiple of 9

but if, n-1 is even then n+1 is also even, then it is a multiple of 4

so, largest multiple = max(9*2, 4*3) = max(18, 12) = 18