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

Explanation:

Let the smaller integer be x, then the larger integer is x + 3.

The sum of the reciprocal of the smaller and twice the reciprocal of the larger is 9/10:

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

Expanding both sides:

10/x + 20/(x + 3) = 90/10

Combining like terms on the left side:

(10 + 20)/(x(x + 3)) = 90/10

30/(x(x + 3)) = 9/10

Cross multiplying both sides:

30 * 10 = 9 * (x(x + 3))

300 = 9x(x + 3)

Expanding the right side:

300 = 9x^2 + 27x

Subtracting 27x from both sides:

273 = 9x^2 + 27x - 27x

273 = 9x^2

Taking the square root of both sides:

√273 = √(9x^2)

√273 = 3x

Dividing both sides by 3:

√273/3 = x

The smaller integer is √273/3 and the larger integer is √273/3 + 3.