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.
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