Re: Inverse Function
- To: mathgroup at smc.vnet.net
- Subject: [mg123629] Re: Inverse Function
- From: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>
- Date: Tue, 13 Dec 2011 05:44:09 -0500 (EST)
- Delivered-to: l-mathgroup@mail-archive0.wolfram.com
- References: <jc4plc$ckp$1@smc.vnet.net>
On Mon, 12 Dec 2011 11:47:56 -0000, Harry Har <harryhar800 at gmail.com>
wrote:
> Hi, All,
>
> I'm newbie in Mathematica. I want to find the inverse function of y=(ax
> +b)/(cx+d). How to to this in Mathematica? Many thank's.
>
> Harry.
>
Reduce[y == (a x + b)/(c x + d), x]
gives
(b == d y && a == c y && d + c x != 0) ||
(a - c y != 0 && x == (-b + d y)/(a - c y) && b c - a d != 0)
but if you don't care about the conditions you can use
Solve[y == (a x + b)/(c x + d), x]
giving
{{x -> (-b + d y)/(a - c y)}}