Question

A positive integer is 3 less than another. If the sum of the reciprocal of the smaller and

twice the reciprocal of the larger is 9/10 then find the two integer

Answer 1

Let's call the smaller integer "x". Then the larger integer would be "x + 3".

The sum of the reciprocal of the smaller integer and twice the reciprocal of the larger integer is 9/10, so we can write that as:

1/x + 2/(x + 3) = 9/10

Expanding and simplifying the right side:

1/x + 2/(x + 3) = 9/10

(x + 6)/(x * (x + 3)) = 9/10

10(x + 6) = 9x * (x + 3)

10x + 60 = 9x^2 + 27x

9x^2 - 83x + 60 = 0

Now we have a quadratic equation that we can solve using the quadratic formula or by factoring. Since the coefficients are integers, we can try factoring:

(3x - 20)(3x - 3) = 0

So either 3x - 20 = 0 or 3x - 3 = 0, which means either x = 20/3 or x = 3. Since x must be a positive integer, the only solution is x = 3.

So the two integers are 3 and 3 + 3 = 6.