The senior classes at high school a and high school b planned separate trips to the indoor climbing gym. the senior class at high school a rented and filled 13 vans and 5 buses with 353 students. high school b rented and filled 9 vans and 5 buses with 309 students. each van and bus carried the same number of students. find the number of students in each van and in each bus.

13v + 5b = 353
9v + 5b = 309multiply by -1

13v + 5b = 353
-9v - 5b = -309 (result of multiplying by -1)
add
4v = 44
v = 44/4
v = 11 <=== a van holds 11 students

9v + 5b = 309
9(11) + 5b = 309
99 + 5b = 309
5b = 309 - 99
5b = 210
b = 210/5
b = 42 <=== a bus holds 42 students

Assume that the number of students in a van is x and that the number of students in a bus is y.

For high school A:
Total number of students = 353 students
number of vans = 13
number of buses = 5 buses
This means that:
13x + 5y = 353 > equation I

For high school B:
Total number of students = 309 students
number of vans = 9 vans
number of buses = 5 buses
This means that:
9x + 5y = 309 
5y = 309 - 9x > equation II

Substitute with equation II in equation I:
13x + 5y = 353 
13x + 309 - 9x = 353
4x = 44 
x = 44/4 = 11
Substitute with the value of x in equation II as follows:
5y = 309 - 9x
5y = 309 - 9(11)
5y = 309 - 99
5y = 210
y = 42

Based on the above calculations:
one van can hold 11 students
one bus can hold 42 students

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