Answer:
There are 24 students and teachers that rode in vans, and there are 385 students and teachers that rode in buses.
Step-by-step explanation:
First we are going to set up a system of equations where:
<em>b= number of buses</em>
<em>v= number of vans</em>
We know from the equation that each bus transported 55 people and that each van transported 12 people and that there was a total of 409 people transported. We also know that the number of buses was 5 more than the number of vans.
Knowing that we can set up our system:
![\left \{ {{55b+12v=409} \atop {b=v+5\\}} \right.](https://tex.z-dn.net/?f=%5Cleft%20%5C%7B%20%7B%7B55b%2B12v%3D409%7D%20%5Catop%20%7Bb%3Dv%2B5%5C%5C%7D%7D%20%5Cright.)
Since we know what <em>b</em> is equal to, we can now use the substitution method to plug the value of <em>b</em> from the bottom equation into the one on the top and solve for variable <em>v.</em>
55(<em>v </em>+5)+12<em>v </em>=409
From there we can distribute 55 into (v+5)
55<em>v </em>+275+12<em>v </em>=409
We can now combine like terms.
67<em>v</em> +275 = 409
From there we subtract 275 from both sides of the equation to isolate the variable.
67<em>v +</em>275 = 409
-275 -275
After doing this, we now divide both sides of the equation by 67 to solve for <em>v.</em>
![\frac{67v}{67} =\frac{134}{67}](https://tex.z-dn.net/?f=%5Cfrac%7B67v%7D%7B67%7D%20%3D%5Cfrac%7B134%7D%7B67%7D)
We now know that <em>v, </em>the number of vans is 2 and can plug that into our bottom equation from our system of equations (
).
![b=2+5\\b=7](https://tex.z-dn.net/?f=b%3D2%2B5%5C%5Cb%3D7)
Knowing that there are 7 buses and 2 vans, we can now plug these numbers into the terms in the top equation of our system. <em>55b </em><em>and </em><em>12v.</em>
![55(7)=385](https://tex.z-dn.net/?f=55%287%29%3D385)
![12(2)= 24\\](https://tex.z-dn.net/?f=12%282%29%3D%2024%5C%5C)
385 people rode buses. 24 people rode vans
Check your equation
409 people went in total. 385 + 24 = 409.