The known condition: `1/x` -`1/y` =4   `(3x+5xy-3y)/(x-3xy-y)` =?

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First, multiply the given condition by the LCD which is xy.

`xy*(1/x - 1/y)=4*xy`

`y-x = 4xy`

Then, plug-in this to

`(3x+5xy-3y)/(x-3xy-y)`

But before substituting, group together the 3x and 3y in the numerator. And in the denominator group x and y too.

`=(5xy + (3x-3y))/(-3xy+(x-y))`

Factor out the GCF...

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First, multiply the given condition by the LCD which is xy.

`xy*(1/x - 1/y)=4*xy`

`y-x = 4xy`

Then, plug-in this to

`(3x+5xy-3y)/(x-3xy-y)`

But before substituting, group together the 3x and 3y in the numerator. And in the denominator group x and y too.

`=(5xy + (3x-3y))/(-3xy+(x-y))`

Factor out the GCF in 3x-3y.

`=(5xy+3(x-y))/(-3xy+(x-y))`

To express (x-y) as (y-x), factor out the negative.

`=(5xy-3(-x+y))/(-3xy-(-x+y))`

`=(5xy-3(y-x))/(-3xy-(y-x))`

Then, plug-in y-x =4xy.

`=(5xy-3(4xy))/(-3xy-4xy)`

`=(5xy-12xy)/(-3xy-4xy)`

`=(-7xy)/(-7xy)`

`=1`

Hence, `(3x+5xy-3y)/(x-3xy-y)=1` .

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