Originally Posted by

**Mukilab** $\displaystyle y=\frac{2pt}{p-t}$

rearrange for t to be subject

I did at first

$\displaystyle t=\frac{-2tp}{yp}$

$\displaystyle \color{blue}y=\frac{2pt}{p-t}$

$\displaystyle \color{blue}y(p-t)=2pt$

$\displaystyle \color{blue}yp-yt=2pt$

$\displaystyle \color{blue}yt-yp=-2pt$

this does not simplify to your first answer

but then that still has t in it so I tried

$\displaystyle \frac{y(p-t)}{2p}=\color{red}t\color{black}\rightarrow\frac{ yp-yt}{t}=2p\rightarrow \color{red}yp-y=2p\color{black}\rightarrow y-y=p=0$

you cannot divide t into only one term of the numerator

$\displaystyle \color{blue}\frac{12-6}{3}=\frac{4(3)-3(2)}{3}=4-2$

which is 6 divided by 3

Also you cannot divide just one term on the left by p

$\displaystyle \color{blue}yp-y=2p$

$\displaystyle \color{blue}y(p-1)=2p$

$\displaystyle \color{blue}yp-y=2p,\ \frac{yp-y}{p}=\frac{2p}{p}=2$

so

$\displaystyle 0=\frac{2pt}{p-t}\rightarrow p-t=2pt \rightarrow t=\frac{-2pt}{p}$

Your last line also contains 2 errors, the first part would be correct if the fraction was 1, if the fraction was 0, 2pt=0, etc

does that help at all?

I feel I've gone terribly wrong, any help please?