# One number is 5 less than another. The sum of their reciprocals is 13. Find the two numbers.

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You should come up with the notation x and y for the given numbers such that:

`x = y - 5`

`1/x + 1/y = 13`

You need to solve for x and y the system of equations `x = y - 5 ` and `1/x + 1/y = 13.`

You should substitute y - 5 for x in the equation `1/x + 1/y = 13 ` such that:

`1/(y-5) + 1/y = 13`

You need to bring the terms to a common denominator such that:

`y + y - 5 = 13y(y-5)`

`2y - 5 = 13y^2 - 65y`

You need to move all terms to one side such that:

`13y^2 - 65y - 2y + 5 = 0`

`13y^2 - 67y + 5 = 0`

You should use quadratic formula such that:

`y_(1,2) = (67+-sqrt(4489 - 260))/26`

`y_(1,2)~~ (67+-65)/26`

`y_1~~ 5.076`

`y_2~~ 1/13`

You may find x such that:

`x_1~~ 5.076 - 5 =gt x_1~~ 0.076`

`x_2~~ 1/13 - 5 =gt x_2~~ -64/13`

**Hence, evaluating the numbers that check the conditions yields `x_1~~ 0.076 ; y_1 ~~ 5.076 ; x_2~~ -64/13 ; y_2 ~~ 1/13` .**

One number is 5 less than another. The sum of their reciprocals is 13. Find the two numbers.

`x = y - 5`

`(1)/(x) + (1)/(y) = 13`

`(x+y)/(xy) = 13`

`(y - 5 + y)/(y(y-5)) = 13`

`(2y - 5)/(y^2 - 5y) = 13`

`13(y^2 - 5y) = 2y - 5`

`13y^2 - 65y = 2y - 5`

`13y^2 - 67y + 5 = 0`

Now that you have a simple quadratic. You can solve for y.