Let the two positive integers be x and y. We are given that their difference is 5, so we can write:
x - y = 5 or y - x = 5.
We are also given that the sum of their reciprocals is 9/14. This can be written as:
1/x + 1/y = 9/14.
Let's assume x > y, so x - y = 5. Then x = y + 5.
Substitute x = y + 5 into the second equation:
1/(y+5) + 1/y = 9/14
Multiply both sides by 14y(y+5) to eliminate the fractions:
14y + 14(y+5) = 9y(y+5)
14y + 14y + 70 = 9y^2 + 45y
28y + 70 = 9y^2 + 45y
Rearrange the equation into a quadratic form:
9y^2 + 17y - 70 = 0
We can solve this quadratic equation using factoring or the quadratic formula. Let's try factoring:
(9y - 28)(y + 2.5) is not correct. Try (9y + 35)(y - 2).
The correct factorization is (9y - 28)(y + 2.5) = 0 doesn't work out nicely. Try another approach.
Let's try to find integers that fit. We can rewrite the reciprocal equation as 1/x + 1/y = 9/14. This means that 1/x = 9/14 - 1/y. Since x and y are positive integers, the values of 1/x and 1/y must be positive. Therefore 9/14 > 1/y so y < 14/9 ≈ 1.55.
However, we also know x and y have a difference of 5. So we can say y is close to x, with their sum reciprocals equal to 9/14.
From the options, try y = 2:
If y = 2, then x = y + 5 = 2 + 5 = 7.
Check the reciprocals: 1/2 + 1/7 = (7 + 2)/14 = 9/14. This is correct.
The two numbers are 2 and 7. Since we are looking for one of the numbers, and 2 is an option, the correct answer is 2.
If y - x = 5, then y = x + 5
1/x + 1/(x+5) = 9/14
14(2x + 5) = 9x^2 + 45x
28x + 70 = 9x^2 + 45x
9x^2 + 17x - 70 = 0
(9x - 28)(x + 2.5) — doesn't factor easily.
y = 2, x = 7 → 1/7 + 1/2 = 9/14
The given numbers are 2 and 7. The question asks for one of the numbers.