Exercise 1 Page 412

Let v be the number of students that fit in a van and c be the number of students that fit in a car. Having 3 vans and 1 car, we can transport 22 students. We should write this as an equation. 3 v+ c=22If the group takes 2 vans and 4 cars, 28 students can be transported. We will use this to write our second equation. 2 v+4 c=28 Combining these, we form a system of equations. 3 v+ c=22 & (I) 2 v+4 c=28 & (II) We can solve this by isolating c in Equation (I) and then use the Substitution Method to solve for v in the second equation.

3v+c=22 & (I) 2v+4c=28 & (II)

SubEqn

(I): LHS-3v=RHS-3v

c=22-3v 2v+4c=28

Substitute

(II): c= 22-3v

c=22-3v 2v+4( 22-3v)=28

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(II): Solve for v

Distr

(II): Distribute 4

cy=22-3v 2v+88-12v=28

SimpTerms

(II): Simplify terms

c=22-3v -10v+88=28

SubEqn

(II): LHS-88=RHS-88

c=22-3v -10v=-60

DivEqn

(II): .LHS /(-10).=.RHS /(-10).

c=22-3v v=6

We do not have to solve the entire system as we were only asked to find how many students can fit in each van. We conclude that 6 students can fit in each van, which corresponds to option C.