The senior classes at High School A and High School B planned separate trips to the state fair. The senior class at High School A rented and filled 13 vans and 6 buses with 278 students. High School B rented and filled 10 vans and 12 buses with 428 students. Every van had the same number of students in it as did the buses. Find the number of students in each van and in each bus.

To find the number of students in each van and bus, we can use system of equations to solve. Our first step is to create one equation for High School A and B. Let's use x for vans and y for buses, but ultimately, the variables don't matter that much.

High School A: 13x+6y=278

High School B: 10x+12y=428

Let's use elimination method to solve. We would multiply the first equation by 2.

26x+12y=556

10x+12y=428

Now that the 12y is the same in both equations, let's eliminate y by subtracting the 2 equations.

26x+12y=556

-(10x+12y=428)

16x=128

With 16x=128 left, we can just solve for x.

16x=128 [divide both sides by 16]

x=8

Now that we have x=8, we can plug that into the equations above to find y.

13(8)+6y=278 [multiply]

104+6y=278 [subtract both sides by 104]

6y=174 [divide both sides by 6]

y=29

With x=8 and y=29, we know that there are 8 students in each van and 29 students in each bus.

Аноним Аноним · Answer 2 · 2021-12-20T07:21:43+01:00

Аноним Аноним

20-12-2021

Step-by-step explanation:

Let,

No. of students in vans = x

No. of students in bus = y

In case 1

13x + 6y = 278(1)

In case 2

10x + 12y = 428(2)

On multiplying eq. 1 by 2

2(13x + 6y) = 2(278)

26x + 12y = 556 (3)

On subtracting 2 and 3

26x + 12y - (10x + 12y) = 556 - 428

26x + 12y - 10x - 12y = 128

16x = 128

x = 128/16

x = 8

From 1

13(8) + 6y = 278

104 + 6y = 278

6y = 278 - 104

6y = 174

y = 174/6

y = 29

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