Two numbers are in the ratio 3 : 5. If 9 is subtracted from the numbers, the…

Concept — A ratio a : b fixes only the relative size of two quantities, never their actual values, so every pair standing in that ratio can be written as a·k and b·k for one common positive multiplier k. A statement that two ratios are equal is a proportion, and a proportion is cleared of fractions by cross-multiplication: from p/q = r/s it follows that p·s = q·r. Determining the actual numbers therefore reduces to finding the single value of k that also satisfies the second condition.

Application — Introduce the common multiplier, then let the second ratio pin it down.

1. Let the two numbers be 3k and 5k, so their ratio is 3 : 5 for every positive value of k.

2. Subtracting 9 from each number gives 3k − 9 and 5k − 9.

3. These are stated to be in the ratio 12 : 23, which is the proportion (3k − 9) / (5k − 9) = 12 / 23.

4. Cross-multiply to clear the fractions: 23(3k − 9) = 12(5k − 9).

5. Expand both sides: 69k − 207 = 60k − 108.

6. Collect the k terms on one side and the constants on the other: 69k − 60k = 207 − 108, that is 9k = 99.

7. Divide both sides by 9: k = 11.

8. Substitute k back into 3k and 5k: the two numbers are 33 and 55.

Cross-check — Dividing 33 : 55 by their common factor 11 gives 3 : 5, so the first condition holds. Subtracting 9 gives 33 − 9 = 24 and 55 − 9 = 46, and dividing 24 : 46 by their common factor 2 gives 12 : 23, so the second condition holds as well.

Why the second condition decides it — Every pair offered here already stands in the ratio 3 : 5, so the first condition alone cannot single one out; only the ratio left after 9 is taken from each number distinguishes them. The same cross-multiplication also gives a quick general test: for two numbers a and b, taking one and the same amount c away from both leaves them in the ratio 12 : 23 exactly when 23(a − c) = 12(b − c), that is when 23a − 12b = 11c.