make " y " the subject?

x = 2 + 3y / y - 5

xy - 5x = 2 + 3y

xy - 3y = 2 + 5x

y ( x - 3 ) = 2 + 5x

y = 2 + 5x / x - 3

x = 2+ 3y / y - 5

1.Add 5 to both sides (they remain equal): x + 5 = 2 + 3y / y

2.Take 2 from both sides: x + 5 - 2 = 3y / y

3.Times both sides by y: (x + 5 - 2) * y = 3y

4. Divide both sidesby 3: ((x + 5 - 2) * y) / 3 = y

:)

(Edited, sorry I made a mistake the first time)

Bring the RHS under a common denominator

x=(2(y-5)+3y)/(y-5)

cross-multiply

x(y-5)=2y-10+3y

bring the 5y from the RHS to the LHS and take y as a common factor

y(x-5)=5(x-2) y=5(x-2)/(x-5)

Multiply both sides of the equation by (y-5) ...

x(y-5) = 2+3y

Multiply out the brackets ...

xy - 5x = 2 + 3y

Subtract 3y from both sides ...

xy - 3y - 5x = 2

Add 5x to both sides ...

xy - 3y = 5x + 2

Factorise the left hand side to isolate y ...

y(x-3) = 5x+2

Divide both sides by (x-3) ...

y = (5x+2) / (x-3)

To be strictly correct brackets must be used.

x = (2 + 3y) / (y - 5)

xy - 5x = 2 + 3y

(x - 3)y = 2 + 5x

y = (2 + 5x) / (x - 3)

Not sure if you wanted this solved, so I'll just make y on the left side...

y-5 = 7 + 3y / x