A positive integer is 2 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 11/15,
then find the two integers.

Step-by-step explanation:

Let x represent the smaller integer.

Let y represent the larger integer.

A positive integer is 2 less than another. It means that

y = x + 2

If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is 11/15, it means that

1/x + 2/y = 11/15- - - - - - - - - - 1

Substituting y = x + 2 into equation 1, it becomes

1/x + 2/(x + 2) = 11/15

Considering the left hand side of the equation, it becomes

(x + 2 + 2x)/x(x + 2)

Therefore,

(3x + 2)/(x² + 2x) = 11/15

Cross multiplying, it becomes

15(3x + 2) = 11(x² + 2x)

45x + 30 = 11x² + 22x

11x² + 22x - 45x - 30 = 0

11x² - 23x - 30 = 0

11x² + 10x - 33x - 30 = 0

x(11x + 10) - 3(11 + 10) =℅0

(x - 3)(11x + 10) = 0

x - 3 = 0 or 11x + 10 = 0

x = 3 or x = - 10/11

Since it us a positive integer, the smaller integer is 3 and the larger integer is 3 + 2 = 5

6 and 24

step-by-step explanation:

6 x 4 = 24