The sum of two natural numbers is 8. Determine the numbers if the sum of their…

For two numbers x and y, the identity 1/x + 1/y = (x + y) / (xy) rewrites a reciprocal-sum condition in terms of their sum and product. This turns a fractional equation into a simple factor-pair search once the sum and product are known.

1. Let the two natural numbers be x and y. Given: x + y = 8 and 1/x + 1/y = 8/15.

2. Using the identity, 1/x + 1/y = (x + y)/(xy), so (x + y)/(xy) = 8/15.

3. Substituting x + y = 8: 8/(xy) = 8/15, which gives xy = 15.

4. So x and y are natural numbers with x + y = 8 and xy = 15. The factor pairs of 15 are (1, 15) and (3, 5); only 3 and 5 add up to 8.

5. Therefore, the two numbers are 3 and 5.

Check: 3 + 5 = 8, and 1/3 + 1/5 = 5/15 + 3/15 = 8/15 — both conditions are satisfied, confirming the numbers 3 and 5.