Comparing two purchases

One cap and three badges cost seventeen dollars and forty cents. Two such caps and nine such badges cost forty-eight dollars and thirty cents. Find the cost of one badge. Try it first. Pause if you like. We need matching cap counts. What happens if we double the first whole purchase? One cap and three badges cost seventeen dollars and forty cents. Two copies of that purchase contain two caps and six badges. Double its whole price too. Two caps and six badges cost two times seventeen dollars and forty cents, which is thirty-four dollars and eighty cents. The other purchase has two caps and nine badges, costing forty-eight dollars and thirty cents. Both rows now contain two caps. The caps have the same cost in each row, so they contribute nothing to the difference. Nine badges minus six badges leaves three extra badges. The difference between the totals pays for those three badges. Three badges cost forty-eight dollars and thirty cents minus thirty-four dollars and eighty cents, which is thirteen dollars and fifty cents. The three badges are identical. Divide their total cost by three to find one badge. One badge costs thirteen dollars and fifty cents divided by three, which is four dollars and fifty cents. Multiply the whole purchase, counts and price together. Match one item, then use the difference to price the other.

One pencil case and two markers cost nine dollars and seventy cents. Four such pencil cases and two such markers cost twenty-two dollars and ninety cents. Find the cost of one pencil case. Try it first. Pause if you like. Which item already has the same count in both purchases? One pencil case and two markers cost nine dollars and seventy cents. Four pencil cases and two markers cost twenty-two dollars and ninety cents. Both purchases contain two markers. Their marker costs already match. We can compare these purchases without scaling. Four pencil cases minus one pencil case leaves three extra pencil cases. The markers make no difference to the total cost. Three pencil cases cost twenty-two dollars and ninety cents minus nine dollars and seventy cents, which is thirteen dollars and twenty cents. One pencil case costs thirteen dollars and twenty cents divided by three, which is four dollars and forty cents. When one item matches, the difference in total cost belongs to the extra items. Divide by their count to find the cost of one.