Which expression is equivalent to the following complex fraction? -2/x+5/y/3/y-2/x

The way how the question entered here, I have to make few assumptions to answer. Hoping that one of the below assumption is your actual problem.Assumption 1: If your complex fraction is given as below[tex] \frac{\frac{-2}{x+\frac{5}{y}}}{\frac{3}{y-\frac{2}{x}}} [/tex]Then we need to write [tex] x+\frac{5}{y} [/tex] as a single fraction by rewriting x as a fraction [tex] \frac{x}{1} [/tex], then multiplying y on both top and bottom of that fraction to get [tex] \frac{xy}{y} [/tex].We can rewrite [tex] x+\frac{5}{y} [/tex] as [tex] \frac{xy}{y} +\frac{5}{y} [/tex] which will be equal to [tex] \frac{xy+5}{y} [/tex]. In the same way[tex] y-\frac{2}{x} =\frac{xy}{x}-\frac{2}{x}=\frac{xy-2}{x} [/tex]When dividing fractions we need to flip the bottom and multiply with top:So [tex] \frac{-2}{\frac{xy+5}{y}} [/tex][tex] \frac{-2}{1} * \frac{y}{xy+5}=\frac{-2y}{xy+5} [/tex]In the same way, [tex] \frac{3}{\frac{xy-2}{x}} =\frac{3}{1} *\frac{x}{xy-2} =\frac{3x}{xy-2} [/tex]Applying the same logic of flipping the fraction and multiplying with the top, we get the below final expression.[tex] \frac{\frac{-2y}{xy+5} }{\frac{3x}{xy-2} } =\frac{-2y}{xy+5}*\frac{xy-2}{3x}=\frac{-2y(xy-2)}{3x(xy+5)}=\frac{4y-2xy^{2}}{15x+3x^{2}y} [/tex]-----Assumption 2: If your complex fraction is given as below[tex] \frac{\frac{-2}{x} +\frac{5}{y} }{\frac{3}{y} -\frac{2}{x} } [/tex]The LCD of both the top part and bottom part is 'xy', we can multiply that in both top and bottom part to simplify the given complex fraction as below:[tex] \frac{xy(\frac{-2}{x}+\frac{5}{y}) }{xy(\frac{3}{y}-\frac{2}{x}) } =\frac{\frac{-2xy}{x}+\frac{5xy}{y}}{\frac{3xy}{y}-\frac{2xy}{x}} [/tex]Cancelling the common factors in both top and bottom of each fraction we will get the below simplified fraction[tex] \frac{-2y+5x}{3x-2y} =\frac{5x-2y}{3x-2y} [/tex]