math question involving inverses!!!?

Make it y=(x +3)/(x-2)

Swap the x and y

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

Now solve for y

x(y-2)=y+3

xy-2x=y+3

xy-y=2x+3

y(x-1)=2x+3

y=(2x+3)/(x-1)

The inverse function of: f(x) = (x +3)/(x-2) is:

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f ˉ¹(x) = (2x + 3) / (x-1)

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There are, in fact, two (2) "y" variables, but there is, in fact, an inverse function. Here is how to find it:

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Given: f(x)= y = (x +3) / (x-2) '

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1) Re-write by switching the "y" variables to "x"; and the "y" variables to "y"

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(By the way, the symbol for the "inverse function" is: f ˉ¹(x) .

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x =(y+3) / (y-2) ---> Now we want to isolate the ["new"] y-value:

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x*(y-2) = y+3 ----? xy - 2x = y + 3 ---> Now, add "2x" to EACH side of the equation; Subtract "y" from each side of the equation: ----> xy - 2x + 2x -y = y + 3 - y + 2x

-----> xy -y = 2x +3 ---> Now , we takes steps to further isolate "y":

Factor out a "y" on the ';left-hand side"---and rewrite:

y(x-1) = 2x +3 ----> Now, we can further isolate the "y", by dividing EACH side of the equation by "(x-1)":

[y(x-1)]/ (x-1) = (2x + 3) / (x-1) ; to get:

y = (2x + 3) / (x-1)

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So, the inverse function of f(x) = (x +3)/(x-2) is:

f ˉ¹(x) = (2x + 3) / (x-1)

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{Note: f ˉ¹(x) is the symbol used to represent the inverse function}.

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Hopefully, this information is of some help to you.

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

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

xy - 2x = y + 3

xy - y = 3 + 2x

y(x - 1) = 3 + 2x

y = (3 + 2x)/(x - 1)

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

xy-2x = y+3

Can you do it from there?