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Topic review (newest first)

Ricky
2005-12-15 06:58:23

Whoops, you used a different way of finding the inverse than I'm used to, so I got mixed up.

What I normally do is:

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

What you did was pretty much the same, but you switched the x and y variables, as it is standard to want to solve for y:

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

As you can see, you come up with the same answer, just different names for the variables.  In mine, x is dependant on y, in yours (as it normally is), y is dependant on x.

student007
2005-12-14 17:51:55

thank you

Ricky
2005-12-14 17:43:23

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

student007
2005-12-14 17:39:45

Find inverse of this function
F(X) = X/3X-2

Interchange X & Y
Y = X/3X-2
X = Y/3Y-2
X(3Y-2) = Y
3XY-2X = Y

I'm not sure what to do from here any help would be appreciated. Thanks

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