Answer:
Let's first define our variables:
Let x be the number of students in each van and bus.
Using substitution:
From the problem, we know that:
High School A rented 8 vans and 8 buses, with a total of 240 students. So we can write the equation:
8x + 8x = 240
High School B rented 4 vans and 1 bus, with a total of 54 students. So we can write the equation:
4x + 1x = 54
Now, we can solve for x in one of the equations and then substitute that value into the other equation to solve for the other variable. For example, let's solve for x in the second equation:
5x = 54
x = 10.8
Now, we can substitute this value of x into the first equation to solve for the number of students in each van and bus for High School A:
8x + 8x = 240
8(10.8) + 8(10.8) = 172.8
So each van and bus for High School A has 10.8 students in it.
Using elimination:
We can rewrite the equations we used above in standard form:
8x + 8y = 240
4x + y = 54
We can eliminate y by multiplying the second equation by -8 and adding it to the first equation:
8x + 8y = 240
-32x - 8y = -432
-24x = -192
x = 8
Now, we can substitute this value of x into either equation to solve for y:
4(8) + y = 54
y = 22
So each van and bus for High School A has 8 students in it and each van and bus for High School B has 22 students in it.
Using graphing:
We can graph the two equations on the same coordinate plane and find the point where they intersect, which represents the solution:
8x + 8y = 240
4x + y = 54
To graph these equations, we can first solve for y in each equation:
y = -x + 30
y = -4x + 54
Then, we can plot these two lines on the same coordinate plane and find their intersection:
(6, 24)
So each van and bus for High School A has 6 students in it and each van and bus for High School B has 24 students in it.