You need to bring the terms in equation 1 to a common denominator, such that:

`1/x + 1/y = 1 => (y + x)/(xy) = 1 => y + x = xy`

You need to bring the terms in equation 2 to a common denominator, such that:

`(3y + x)/(xy) = 2 => 3y + x = 2xy`

You need to substract the equation `y + x = xy ` from `3y + x = 2xy` , such that:

`3y + x - y - x = 2xy - xy `

`2y = xy => 2y - xy = 0 `

Factoring out y yields:

`y(2 - x) = 0 => y = 0 or 2 - x = 0 => y = 0, x = 2`

Substituting 2 for x in the first equation yields:

`2 + y = 2y => y - 2y = -2 => -y = -2 => y = 2`

**Hence, testing if there exists a solution to the system of equations yields **`x = 2, y = 2.`

1/x + 1/y = 1

x + y = xy (1)

3/x + 1/y = 2

3y + x = 2xy (2)

We'll replace (1) in (2):

3y + x = 2(x+y)

3y + x = 2x + 2y

3y - 2y = 2x - x

y = x (3)

We'll replace (3) in (2)

3x + x = 2x^2

4x = 2x^2

We'll divide by 2 and we'll use the symmetric property:

x^2 - 2x = 0

We'll factorize by x:

x(x-2) = 0

x = 0

x-2 = 0

x = 2

Since x = y => y = 2

The solution of the system is represented by the pair of coordinates (2 , 2).