Transposition

**andyw** · June 22nd 2009, 01:19 PM

Hi could someone please tell me if this is correct, and steer me in the right direction if it isn't?

Transpose the following for r:

$x=\frac{1}{r-1}-\frac{1}{r+1}+1$

$x-1=\frac{1}{r-1}-\frac{1}{r+1}$

$r-1=\frac{1}{x-1}-\frac{1}{r+1}$

$(r-1)(r+1)=\frac{1}{x-1}-1$

$r^2-1=\frac{1}{x-1}-1$

$r^2=\frac{1}{x-1}$

$r=\sqrt\frac{1}{x-1}$

Thanks in advance!

**yeongil** · June 22nd 2009, 01:40 PM

Originally Posted by andyw

$x=\frac{1}{r-1}-\frac{1}{r+1}+1$
$x-1=\frac{1}{r-1}-\frac{1}{r+1}$
$r-1=\frac{1}{x-1}-\frac{1}{r+1}\;\;\;{\color{red}This\;is\;incorrect ...}$
$(r-1)(r+1)=\frac{1}{x-1}-1\;\;\;{\color{red}And\;so\;is\;this...}$
...

You want to solve for r, right?
$\begin{aligned} x &= \frac{1}{r - 1} - \frac{1}{r + 1} + 1 \\ x - 1 &= \frac{1}{r - 1} - \frac{1}{r + 1} \\ (x - 1){\color{red}(r^2 - 1)} &= \left(\frac{1}{r - 1} - \frac{1}{r + 1}\right){\color{red}(r^2 - 1)} \\ (x - 1)(r^2 - 1) &= r + 1 - (r - 1) \\ xr^2 - x - r^2 + 1 &= r + 1 - r + 1 \end{aligned}$

$\begin{aligned} xr^2 - r^2 - x + 1 &= 2 \\ xr^2 - r^2 &= x + 1 \\ r^2(x - 1) &= x + 1 \\ r^2 &= \frac{x + 1}{x - 1} \\ r &= \sqrt{\frac{x + 1}{x - 1}} \\ \end{aligned}$

01

**Bruno J.** · June 22nd 2009, 01:41 PM

$x=\frac{1}{r-1}-\frac{1}{r+1}+1 $

$= \frac{r+1-(r-1)+r^2-1}{r^2-1} = \frac{r^2+1}{r^2-1}$

$x(r^2-1)= r^2+1$

$r^2(x-1)=x+1$

$r=\sqrt{\frac{x+1}{x-1}}$