Set up a system of equations then solve.

Question

The senior classes of High school A and High school B planned separate trips to the indoor climbing gym. The senior class of High school A rented and filled 2 vans and 3 buses with 105 students. High school B rented and filled 14 vans and 6 buses with 270 students. Each van and each bus carried the same amount of students. How many students can a van carry? How many students can a bus carry?

Answer 1

First, define your variables.

x: # of students that can fit in a van

y: # of students that can fit in a bus

Then you need to create the system of equations. The buses have been filled, so we don't need to worry about them being only partially filled. To know the total number of students, we would multiply the number of students on buses by the the number of buses, multiply the number of students on vans by the number of vans, and add the 2 products. With this logic, we can create the system:

  2x+3y=105

14x+6y=270

There are 3 methods to solving a system of equations: elimination, substitution, and graphing.

Remember that in algebra we can do anything we want to an equation as long as we do it to both sides. If we used the elimination method, we would multiply one of the equations by a number that would allow us to add the equations and cancel out one of the variables. The top equation has smaller coefficients and it seems like they would be easier to change into coefficients where one could cancel when added to the corresponding coefficient of the other equation. I will multiply the top one by -2 to get

-4x-6y=-210

Now to combine these equations and solve for the remaining variable

14x+6y=270

-4x - 6y=-210

_____________

 

10x  =  60

    x  =  6

 

Another way to think about the elimination method is that we are merely adding an expression to both sides of one equation that would happen to make the equation solveable

14x+6y=270

    -4x-6y=-4x-6y

14x+6y+(-4x-6y)=270+(-4x-6y)

 

and then we merely use the fact that -4x-6y=-210 to change the expression on the right into a single number

14x+6y-4x-6y=270+-210

 

We can now plug this into one of the equations and solve for the remaining variable y.

 

2(6)+3y=105

12  +3y=105

        3y=93

          y=31

 

Using the substitution method, we would put one of the equations into the form of x= or y= so that we can make a substitution into the other equation. Again, the one with smaller coefficients seems like it would be easier to work with

2x+3y=105

      2x=105-3y

        x=(105-3y)/2

 

We would then plug this into the other equation

14x+6y=270

14(105-3y)/2 + 6y = 270

7(105-3y)+6y=270

   735-21y+6y=270

         735-15y=270

               -15y=-465

                    y=31

 

As we did after using elimination, we plug our known variable into one of the original equations to find the other.

 

    2x+3y=105

2x+3(31)=105

    2x+93=105

          2x=12

            x=6

 

Thirdly, we could graph. We would need to put these 2 equations into a slope-intercept form to easily graph them, which involves putting y by itself on one side.

2x+3y=105

3y=105-2x

y=(105-2x)/3

y=35-(2/3)x

 

14x+6y=270

6y=270-14x

y=(270-14x)/6

y=45-(14/6)x

y=45-(7/3)x

 

We would then graph these 2 lines and then find the point at which they intersect. Note that you could also take these 2 equations and set the right sides equal to each other and solve as a form of substitution. 

 

 

A van can hold 6 students and a bus can hold 31.  I hope this helped!