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Math Help - Rearraging the formula

  1. #1
    Senior Member Mukilab's Avatar
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    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?
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  2. #2
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    y = \frac{2pt}{p-t}

    yp-yt = 2pt<br />

    yp = 2pt+yt

    yp = t(2p+y)

    \frac{yp}{2p+y} = t
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  3. #3
    A riddle wrapped in an enigma
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    Quote Originally Posted by Mukilab View Post
    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..
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  4. #4
    Senior Member Mukilab's Avatar
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    heh, always seem to forget factorising
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  5. #5
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    Quote Originally Posted by Mukilab View Post
    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}
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  6. #6
    Senior Member Mukilab's Avatar
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    Thanks but I didn't explain when I said that I forgot to factorise it

    I had written down this:

    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.
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