# Rearraging the formula

• May 31st 2010, 10:11 AM
Mukilab
Rearraging the formula
$y=\frac{2pt}{p-t}$

rearrange for t to be subject

I did at first
$t=\frac{-2tp}{yp}$

but then that still has t in it so I tried
$\frac{y(p-t)}{2p}=2\rightarrow\frac{yp-yt}{t}=2p\rightarrow yp-y=2p\rightarrow y-y=p=0$

so

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

does that help at all?

I feel I've gone terribly wrong, any help please?
• May 31st 2010, 10:19 AM
skeeter
$y = \frac{2pt}{p-t}$

$yp-yt = 2pt
$

$yp = 2pt+yt$

$yp = t(2p+y)$

$\frac{yp}{2p+y} = t$
• May 31st 2010, 10:20 AM
masters
Quote:

Originally Posted by Mukilab
$y=\frac{2pt}{p-t}$

rearrange for t to be subject

I did at first
$t=\frac{-2tp}{yp}$

but then that still has t in it so I tried
$\frac{y(p-t)}{2p}=2\rightarrow\frac{yp-yt}{t}=2p\rightarrow yp-y=2p\rightarrow y-y=p=0$

so

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

does that help at all?

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

Hi Mukilab,

You'll want to get the terms that have t in them together on one side of the equation.

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

$y(p-t)=2pt$

$py-ty=2pt$

$py=2pt+ty$

$py=t(2p+y)$

$\frac{py}{2p+y}=t$

EDIT: Less than a minure too late..
• May 31st 2010, 10:40 AM
Mukilab
heh, always seem to forget factorising
• May 31st 2010, 11:03 AM
Quote:

Originally Posted by Mukilab
$y=\frac{2pt}{p-t}$

rearrange for t to be subject

I did at first
$t=\frac{-2tp}{yp}$

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

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

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

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

this does not simplify to your first answer

but then that still has t in it so I tried
$\frac{y(p-t)}{2p}=\color{red}t\color{black}\rightarrow\frac{ yp-yt}{t}=2p\rightarrow \color{red}yp-y=2p\color{black}\rightarrow y-y=p=0$

you cannot divide t into only one term of the numerator

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

which is 6 divided by 3

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

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

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

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

so

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

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?

Hi Mukilab,

you have quite a bit of work to do to master this....

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

When dealing with a fraction, multiply both sides of the equality by the denominator,

just as in $3=\frac{6}{2}\Rightarrow\ 3(2)=\frac{6(2)}{2}=(6)\left(\frac{2}{2}\right)=6( 1)=6$

Hence, the 1st step is

$\frac{2pt(p-t)}{p-t}=y(p-t)$

$2pt=y(p-t)$

You want t by itself, but it's in two places here,
not only that but it's on opposite sides.
Hence you next bring the 2 t's to the same side, as in 3x=x+10, 3x-x=x-x+10, 2x=10

Before we bring the t's together, we need to multiply out the right hand side.

2(4+5)=2(9)=18
2(4+5)=2(4)+2(5)=8+10=18

$y(p-t)=yp+y(-t)=yp-yt$

therefore, we have

$2pt=yp-yt$

We get the t's on the same side by adding yt to both sides as we will then have no t's on the right

$2pt+yt=yp-yt+yt$

$2pt+yt=yp$

Now we factorise the left to have t only once

$t(2p+y)=yp$

this is now the same as $2(3)=6$ divide both sides by 3... $\Rightarrow\ 2=\frac{6}{3}$

hence divide both sides by (2p+y)

$t=\frac{yp}{2p+y}$
• May 31st 2010, 11:32 AM
Mukilab
Thanks but I didn't explain when I said that I forgot to factorise it

I had written down this:http://www.mathhelpforum.com/math-he...938c8a94-1.gif

but I didn't think it was of any significance because I forgot to factorise it so I scrapped it in favour of that new, incorrect way.

Thanks for taking the time to explain it the other way though.