Need help finding the inverse of this equation

[stumped]

y = (3x+2)/(2x-5)

I switched the variables and tried to solve for y, however, the fraction part keeps stumping me.

What I've done so far:

x=(3y+2)/(2y-5)
x(2y-5)=3y+2
x(2y-5)-2=3y
[x(2y-5)-2]/3=y

Is this correct so far, and how do I proceed?

 

[Daniel_Feldman]

Stumped-

Here's what you did...

x=(3y+2)/(2y-5) Correct
x(2y-5)=3y+2 Correct
x(2y-5)-2=3y Here's where you make life complicated...
[x(2y-5)-2]/3=y

Remember that you need y on one side and all else on the other side.



Let's check this out.....

y = (3x+2)/(2x-5)

x=(3y+2)/(2y-5)

2yx-5x=3y+3 Same thing you did, but I multiplied it out x(2y-5)=2yx-5x

2yx-3y=5x+2 Ok, all y's on one side. Now I need to move x to the other side. See how to do it?

y(2x-3)=5x+2

y=(5x+2)/(2x-3)


Cheers,

-Daniel-

 

[stumped]

Thank you Sometimes the KISS (Keep It Simple Stupid) method eludes me.

 

[Denis]

Thank you Sometimes the KISS (Keep It Simple Stupid) method eludes me.

Keep It Simple Stumped !