SenorMathWiz
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The senior classes at High School A and High School B planned separate trips to Yellowstone National Park. The senior class at High School A rented and filled 12 vans and 11 buses with 576 students. High School B rented and filled 4 vans and 9 buses with 384 students. Each van and each bus carried the same number of students. How many students rode in each van?

Respuesta :

anna2023147

Answer:

15 students

Step-by-step explanation:

To solve this question, we can set up a system of equations.

Let's start by naming our variables.

We can call the number of students that rode in a van "x".

Then, let's name the number of students that rode a bus "y".

Knowing this, we can create two equations and then solve.

For High School A, the equation would be as follows:

12x+11y=576

For High School B, it would be:

4x+9y=384

What you have to do now, is to eliminate one of the variables.

I'm going to choose to do it with elimination, but substitution also works.

I'm going to multiply the second equation by 3, so that we can subtract the equations to eliminate the x variable.

3(4x+9y=384)

Since we're multiplying 3 to both sides, the equation will remain true.

12x+27y=1152

Now, let's subtract the other equation from this one:

  12x+27y=1152

-  12x+11y=576

_____________

0x+16y=576

Our new equation is:

16y=576

Divide both sides by 16 to isolate y

y=36

This means that 36 students rode in each bus.

Let's now use this info to solve for x.

We can do this by plugging in 36 for y.

4x+9y=384

4x+9(36)=384

Simplify

4x+324=384

Subtract 324 from both sides

4x=60

Divide both sides by 4

x=15

This means that 15 students rode in each van.

As I mentioned earlier, this could have also been solved by substitution.

You would have had to isolate one variable so that one of the equations read x= something or y= something

After that, you plug that in to the other equation so you're left with only one variable.

You then simplify, and plug the value back in for the other variable.

Given the nature of this question, that would have involved working with decimals, so I chose elimination instead.

Please let me know if you have any questions!

Starrysoul100

[tex]\bold{\huge{\underline{ Solution }}}[/tex]

Here, we have given that,

  • The senior classes at High school A and High school B planned separated trips.
  • The number of bus and vans rented by high school class A are 12 vans and 11 buses.
  • The number of bus and vans rented by high school class B are 4 vans and 9 buses.
  • The total number of students in class A are 576 students.
  • The total number of students in class B are 384 students .

Let the number of students in van be 'x' and number of students in bus be 'y' .

According to the question,

Linear equation for class A

[tex]\sf{ 12x + 11y = 576 ...eq(1)}[/tex]

Linear equation for class B

[tex]\sf{ 4x + 9y = 384 ...eq(2) }[/tex]

Subtract eq(2) from eq(1) :-

[tex]\sf{ 12x + 11y -( 4x + 9y) = 576 - 384 }[/tex]

[tex]\sf{ 12x + 11y - 4x - 9y = 192 }[/tex]

[tex]\sf{ 8x + 2y = 192 }[/tex]

[tex]\sf{ 2(4x + y) = 192 }[/tex]

[tex]\sf{ 4x + y = {\dfrac{192}{2}}}[/tex]

[tex]\sf{ 4x + y = 96}[/tex]

[tex]\bold{ y = 96 - 4x ...eq(3) }[/tex]

Subsitute eq (3) in eq(1) :-

[tex]\sf{ 12x + 11( 96 - 4x) = 576 }[/tex]

[tex]\sf{ 12x + 1056 - 44x = 576 }[/tex]

[tex]\sf{ 12x - 44x = 576 - 1056}[/tex]

[tex]\sf{ -32x = - 480 }[/tex]

[tex]\sf{ x = {\dfrac{-480}{-32}}}[/tex]

[tex]\bold{ x = 15 }[/tex]

Thus, The value of x is 15 .

Now, subsitute the value of x in eq(2) :-

[tex]\sf{ 4(15) + 9y = 384 }[/tex]

[tex]\sf{ 60 + 9y = 384 }[/tex]

[tex]\sf{ 9y = 384 - 60 }[/tex]

[tex]\sf{ 9y = 324 }[/tex]

[tex]\sf{ y = {\dfrac{324}{9}}}[/tex]

[tex]\bold{ y = 36 }[/tex]

Thus, The value of y is 36

  • In this question, we have taken x and y are the number of students in each van and a bus.

Therefore,

The number of students in each van = 15

The number of students in each bus = 36

Hence, The number of students rode in each van are 15 .

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