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A positive integer is 4 less than another. If the sum of the reciprocal of the smaller and twice the reciprocal of the larger is frac(5,6), then find the two integers.

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Let the smaller number is x.

So the bigger number is x+4.

Reciprocal of smaller number is
(1)/(x) and reciprocal of larger number is
(1)/(x+4)

SO our equation is

(1)/(x) + 2* (1)/(x+4) = (5)/(6)

(1)/(x) + (2)/(x+4) = (5)/(6)

(x+4 + 2x)/(x(x+4)) = (5)/(6)

(3x+4)/(x(x+4)) = (5)/(6)

by cross multiplication rule

6(3x+4) = 5(x(x+4))

18x+ 24= 5x^2 + 20x

5x^2 + 20x - 18x - 24 = 0

5x^2 +2x - 24 = 0

Now we can solve it by factorization.

As we can see -24*5 = -120.

So we can write -120 as 12 * (-10) = -120 because 12 - 10 = 2

SO we can write expression as

5x^2 + 12x - 10x - 24 = 0

x(5x+12) - 2(5x+12) = 0

(5x+12)(x-2) = 0
So x = 2 or x =
(-12)/(5)

But we have to choose positive integer. SO we will choose x = 2.

SO the smaller number is 2 and bigger number is 6.
User Maleen Abewardana
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