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:

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

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

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

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

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

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

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

2. Originally Posted by andyw
$\displaystyle x=\frac{1}{r-1}-\frac{1}{r+1}+1$
$\displaystyle x-1=\frac{1}{r-1}-\frac{1}{r+1}$
$\displaystyle r-1=\frac{1}{x-1}-\frac{1}{r+1}\;\;\;{\color{red}This\;is\;incorrect ...}$
$\displaystyle (r-1)(r+1)=\frac{1}{x-1}-1\;\;\;{\color{red}And\;so\;is\;this...}$
...
You want to solve for r, right?
\displaystyle \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}

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

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

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

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

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

4. Originally Posted by yeongil
You want to solve for r, right?
$\displaystyle xr^2 - r^2 - x + 1 = 2$
$\displaystyle xr^2 - r^2 = x - 1$
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. $\displaystyle r=\pm\sqrt{\frac{x+1}{x-1}}$