1. ## Transposition

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:

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

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

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

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

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

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

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

2. 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

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

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

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

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

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

4. Originally Posted by yeongil
You want to solve for r, right?
$
xr^2 - r^2 - x + 1 = 2
$

$
xr^2 - r^2 = x - 1
$
this step is incorrect.

There was no equation to solve by the way. Solution to what? If I set r=2 then x=5/3. Isn't that a solution?

5. Caught it just in time!

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