# Two numbers differ by 3. The sum of their reciprocals is 7/10; find the numbers

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### 1 Answer

Since the two number differ by d, we assume the two numbers to be x and x+d .

The sum of their reciprocals = 7, given.

Therefore multiply by 10x(x+3) both sides of 1/x+1/(x+3) = 7/10:

10x(x+3)/x +10x(x+3)/(x+3) = 7x(x+3).

10(x+3) +10x = 7x(x+3).

20x+30 = 7x^2+21x.

Subtract 20x+30 from both sides:

0 = 7x^2+21x-20x-30 = 0

7x^2+x-30 = 0

7x^2+15x -14x-30 = 0

x(7x+15) - 2(7x+15) = 0.

(x-2)(7x+15) = 0.

x-2 = 0, or 7x+15.

x-2 = 0 gives x = 2 .

7x+15 = 0 gives x = -15/7.

Therefore x = 2 . Or x = -15/7.

Therefore x = 2 and x+3 = 2+3 = 5 are a solution pair. Tally: The sum the reciprocals of 2 and 5 = 1/2+1/5 = (5+2)/10 = 7/10.

Similarly x = -15/7 and x+3 = -15/7+3 = 6/7 are also a pair of solution, as 1/x+1/x+3 = -7/15+7/6 = (-14+35)/30 = 21/30 = 7/10.

So (2, 5) and (-15/7 , 6/7) are solutions.