Trouble understanding rewriting of an expression

I'm trying to inverse a simple function but got stuck with some algebra I cant remember.

I've got f (x) = 2x-1/x -1 and so far rewritten it to y(x-1) = 2x-1 but got stuck here. Reading the solution it show the next step as (y-2)x=y-1 but I'm at a loss trying to understand how they arrived here.(Headbang)

Re: Trouble understanding rewriting of an expression

Quote:

Originally Posted by

**dipsy34** I'm trying to inverse a simple function but got stuck with some algebra I cant remember.

I've got f (x) = 2x-1/x -1

The first think you need to recall is the order of operations.

Re: Trouble understanding rewriting of an expression

Quote:

Originally Posted by

**dipsy34** I'm trying to inverse a simple function but got stuck with some algebra I cant remember.

I've got f (x) = 2x-1/x -1 and so far rewritten it to y(x-1) = 2x-1 but got stuck here. Reading the solution it show the next step as (y-2)x=y-1 but I'm at a loss trying to understand how they arrived here.(Headbang)

I assume that you mean

$\displaystyle y=\frac{2x-1}{x-1}$ ............. If so:

$\displaystyle \begin{array}{rcl}y(x-1)&=&2x-1 \\ yx-y&=&2x-1 \\ 1-y&=& 2x-yx \\ 1-y&=&(2-y) x \\ \frac{1-y}{2-y}&=&x \end{array}$

Re: Trouble understanding rewriting of an expression

Quote:

Originally Posted by

**earboth** I assume that you mean

$\displaystyle y=\frac{2x-1}{x-1}$ ............. If so:

$\displaystyle \begin{array}{rcl}y(x-1)&=&2x-1 \\ yx-y&=&2x-1 \\ 1-y&=& 2x-yx \\ 1-y&=&(2-y) x \\ \frac{1-y}{2-y}&=&x \end{array}$

Super, thanks.