Count a few cases first
At a fair, entry costs $3, and every ride costs $p. We want one rule that gives the cost of any number of rides.
Start with cases small enough to count. One ride costs p + 3: one ride, and the entry. Two rides cost 2p + 3, because there are two rides but still only one entry. Three rides cost 3p + 3. Each extra ride adds one more p to the bill, and the 3 stays exactly as it was.
Every bill has one gold block for the $3 entry, and one p block for each ride.
Letters for the numbers that change
Now let n stand for the number of rides and C for the cost in dollars. With n rides there are n blocks of p, which is n × p, written np. The entry is added once. So C = np + 3.
Say what each letter stands for, with its unit, before you use the formula: n is the number of rides, p is the price of one ride in dollars, and C is the total cost in dollars. This one formula answers every version of the question: 4 rides, 10 rides, or 50 rides.
A formula is written from the structure of the story, not from its numbers. Ask two questions. What is charged again and again? That part is multiplied by the count. What is charged only once? That part is added on its own.
The same shape in another story
You have $10 saved and you add $s every week. After 1 week you have 10 + s, after 2 weeks 10 + 2s, and after w weeks 10 + ws. So the total is T = ws + 10.
This is the same shape as C = np + 3. The amount that repeats, s, is multiplied by the count, w, and the amount that happens once, 10, is added on. A taxi fare with a fixed charge to start and a price for each mile has the same shape too.
Check with a case you can count
Test the formula on a case you can work out without it. Take 2 rides at $5 each. Counting gives 2 × $5 = $10 for the rides, and $3 for the entry, so $13 altogether.
Now put n = 2 and p = 5 into the formula: C = 2 × 5 + 3 = 13. The formula agrees with the count, so it is right for this case.
With the price at $5, the number of rides goes into the box for n: 2 × 5 + 3 = 13.
The usual mistakes
Swapping the two charges: C = 3n + p. That charges the entry on every ride and the ride price only once. The count checks it: 2 rides at $5 would cost 3 × 2 + 5 = 11, not 13.
Multiplying the entry too: C = (p + 3)n. That charges $3 again for every ride. For 2 rides at $5 it gives (5 + 3) × 2 = 16, not 13.
Worked example: A Mobile Top-Up Rule Written as a Formula and Inverted
Question A mobile company has this rule for a top-up: a handling fee of $2 is taken off the amount paid, and every dollar that remains buys 5 minutes of calls. (a) Write a formula for the number of minutes, m, bought by a top-up of p dollars, and find the minutes bought by a top-up of $20. (b) Make p the subject of the formula, and find the top-up that buys 150 minutes.
1.A top-up of p dollars leaves p − 2 dollars after the fee. Each of those dollars buys 5 minutes, so m = 5(p − 2).
The fee leaves p − 2 dollars, and each of them buys 5 minutes: m = 5(p − 2). 2.(a) The formula is m = 5(p − 2). For p = 20, m = 5 × (20 − 2) = 5 × 18 = 90 minutes.
(a) m = 5(p − 2). For p = 20, m = 5 × 18 = 90 minutes. 3.To make p the subject, divide both sides by 5 so that the bracket stands alone: m5 = p − 2.
Divide both sides by 5, so that the bracket stands alone: m5 = p − 2. 4.Add 2 to both sides: m5 + 2 = p, so p = m5 + 2.
Add 2 to both sides: p = m5 + 2. 5.(b) For m = 150, p = 1505 + 2 = 30 + 2 = 32. The top-up is $32. Check: 5 × (32 − 2) = 5 × 30 = 150 minutes.
(b) For m = 150, p = 1505 + 2 = 32, so the top-up is $32.
Answer: (a) m = 5(p − 2), 90 minutes; (b) p = m5 + 2, $32
Common mistakes
- Writing m = 5p − 2. That takes 2 minutes off at the end. The rule takes $2 off the payment before multiplying by 5, so the bracket is needed: m = 5(p − 2).
- Inverting the formula as p = m − 25. The operations are undone in the reverse order: the last thing done was multiplying by 5, so divide by 5 first and then add 2.
More equations and inequalities problems, worked step by step →