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A positive integer is 15 less than another. If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal of the larger integer, then the result is − 3/4 . Find all pairs of integers that satisfy this condition.

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A positive integer is 15 less than another -----> x = y - 15

If 4 times the reciprocal of the smaller integer is subtracted from the reciprocal of the larger integer, then the result is − 3/4
------> (1/y) - (4/x) = -3/4

Use substitution, replace 'x' with 'y-15' in 2nd equation


(1)/(y) - (4)/(y-15) = -(3)/(4) \\ \\ ( (y-15)-4y)/(y(y-15))=-(3)/(4) \\ \\ 4(-3y-15) = -3y(y-15) \\ \\ -12y -60 = -3y^2+45y \\ \\ 3y^2-57y-60 = 0 \\ \\ y^2-19y -20 =0 \\ \\ (y-20)(y+1) = 0 \\ \\ y = 20,y=-1

We know that both x and y must be positive integers, so y =-1 is not a solution.

If y=20, then x = 20-15 = 5

Final answer:
The pair of integers is (5,20)
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