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` .**

## We’ll help your grades soar

Start your 48-hour free trial and unlock all the summaries, Q&A, and analyses you need to get better grades now.

- 30,000+ book summaries
- 20% study tools discount
- Ad-free content
- PDF downloads
- 300,000+ answers
- 5-star customer support

Already a member? Log in here.

Are you a teacher? Sign up now