Inverse function for 2x - 7 / 3x - 2 = y

[WiccaJew]

Hey i need help with finding the inverse of 2x-7/3x-2=y. I tried to follow the book intructions but when i did y(3x-2)=2x-7 then 3xy-2y=2x-7 i got lost. I was told to put the varible x on the same side so i did. 3xy-2x=2y-7 and got 3x(y-2/3)=2y-7

 

[pka]

Re: Inverse function

Start with \(\displaystyle \frac{{2x - 7}}{{3x - 1}} = y\).
Change variables: \(\displaystyle \frac{{2y - 7}}{{3y - 1}} = x\).
Now solve for y in terms of x.

 

[WiccaJew]

Re: Inverse function for 2x - 7 / 3x - 1 = y

I tried im still lost i moved to 2x-7=y(3x-2) then 2x-7= 3xy-2x then i didnt know where to go

 

[skeeter]

\(\displaystyle y = \frac{2x-7}{3x-2}\)

swap variables ...

\(\displaystyle x = \frac{2y-7}{3y-2}\)

multiply both sides by (3y - 2) ...

\(\displaystyle x(3y-2) = 2y-7\)

distribute the left side ...

\(\displaystyle 3xy - 2x = 2y - 7\)

get the terms with y on the same side ...

\(\displaystyle 3xy - 2y = 2x - 7\)

factor out y from the two terms on the left ...

\(\displaystyle y(3x - 2) = 2x-7\)

divide both sides by (3x - 2) ...

\(\displaystyle y = \frac{2x-7}{3x-2}\)

note that you get what you started with ... what does this tell you about the original function and its inverse?

 

[WiccaJew]

hmm my answer sheet says x-7/3x-2

 

[skeeter]

... well, you'll have to decide which of us is correct, won't you?

 

[masters]

... well, you'll have to decide which of us is correct, won't you?

The function got changed somewhere along the way from:

\(\displaystyle y = \frac{2x-7}{3x-2}\)

to

\(\displaystyle y = \frac{2x-7}{3x-1}\)

In which case, the latter's inverse would match wicca's answer sheet.