Forming systems and interpreting solutions in context

Step 1 — Define the variables

Both must be whole numbers and both must be non-negative — noted now, used later.

Step 2 — Part (a): form the equations

Seats — exactly 130 students:

$$11v + 18m = 130 \qquad (1)$$

Cost — the budget:

$$180v + 260m = 2000 \qquad (2)$$

Step 3 — Simplify equation (2)

Every term divides by 20:

$$9v + 13m = 100 \qquad (2')$$

Step 4 — Solve by elimination

Match the $v$ coefficients: $(1) \times 9$ and $(2') \times 11$:

$99v + 162m = 1170 \qquad (3)$ $99v + 143m = 1100 \qquad (4)$

Subtract (4) from (3):

$$19m = 70$$

$$m = \frac{70}{19} \approx 3.68$$

Step 5 — Interpret

Reject the solution with a reason: you cannot hire $3.68$ minibuses.

Step 6 — Answer part (a) properly

The equations demanded exactly 130 seats and exactly $2,000, which is a stricter question than the school actually needs. Reformulating with inequations — at least 130 seats, at most $2,000 — is the right model:

$$11v + 18m \geq 130, \qquad 9v + 13m \leq 100$$

Test some whole-number options:

$v$	$m$	Seats $11v + 18m$	Cost check $9v + 13m$	Verdict
2	6	$22 + 108 = 130$ ✓	$18 + 78 = 96 \leq 100$ ✓	Works
4	5	$44 + 90 = 134$ ✓	$36 + 65 = 101 > 100$ ✗	Over budget
0	8	$144$ ✓	$104 > 100$ ✗	Over budget
7	3	$77 + 54 = 131$ ✓	$63 + 39 = 102 > 100$ ✗	Over budget

Note why the equation version failed but the inequation version succeeded: insisting on spending the budget exactly is an artificial condition the real situation never imposed. Choosing equations where the context wanted inequations is a modelling error, not an arithmetic one.

Step 7 — Part (b): form the new system

Seats — at least 130:

$$11v + 18m \geq 130 \qquad (5)$$

Vehicles — at most 9:

$$v + m \leq 9 \qquad (6)$$

Non-negativity: $v \geq 0$, $m \geq 0$, both whole numbers.

Step 8 — Find the corner of the region

Solve the boundaries $11v + 18m = 130$ and $v + m = 9$ simultaneously.

From the second, $v = 9 - m$. Substitute:

$$11(9 - m) + 18m = 130$$

$$99 - 11m + 18m = 130$$

$$7m = 31 \quad \Rightarrow \quad m = \frac{31}{7} \approx 4.43, \qquad v \approx 4.57$$

Step 9 — List the whole-number combinations

Test each value of $m$ with the smallest $v$ that gives enough seats, then check the vehicle limit:

$m$	Seats from minibuses	Vans needed for 130	$v$	Total vehicles	Verdict
8	144	0	0	8 ✓	Works
7	126	$\lceil 4/11 \rceil = 1$	1	8 ✓	Works
6	108	$\lceil 22/11 \rceil = 2$	2	8 ✓	Works
5	90	$\lceil 40/11 \rceil = 4$	4	9 ✓	Works
4	72	$\lceil 58/11 \rceil = 6$	6	10 ✗	Too many vehicles
3	54	$\lceil 76/11 \rceil = 7$	7	10 ✗	Too many vehicles

Step 10 — State the answer in context

Costs, for the school to choose between:

• 8 minibuses: $2,080 — over the earlier budget
• 7 minibuses + 1 van: $1820 + 180 = \$2{,}000$ — exactly on budget
• 6 minibuses + 2 vans: $1560 + 360 = \$1{,}920$ — cheapest
• 5 minibuses + 4 vans: $1300 + 720 = \$2{,}020$ — over budget

What the model leaves out: driver availability, whether the school has enough licensed drivers for 8 vehicles, and luggage or ski-gear capacity, which may reduce the effective seating well below the stated figures.