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Stuck on yet another maths question - please help watch

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    (Original post by knoxode)
    2r + 2*h = r*h

    -2h

    2r = r - h
    You've fixed your left hand side but you've changed the right hand side, which was correct before. It might be helpful to think about it like this:

     2r+2h=rh 





2r+2h-2h=rh-2h

    Can you see what I've done there and how it simplifies?
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    (Original post by sindyscape62)
    You've fixed your left hand side but you've changed the right hand side, which was correct before. It might be helpful to think about it like this:

     2r+2h=rh 





2r+2h-2h=rh-2h

    Can you see what I've done there and how it simplifies?
    I thought it simplifies too

    2r = rh - 2h

    2r = h(r-2)

    2r/(r-2) = h
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    (Original post by knoxode)
    I thought it simplifies too

    2r = rh - 2h

    2r = h(r-2)

    2r/(r-2) = h
    Yes, that's correct.

    Maybe you could try a similar question to check that you get the method?

     3kx +2y= 2xy

    Can you get x on its own in that equation?
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    (Original post by sindyscape62)
    Yes, that's correct.

    Maybe you could try a similar question to check that you get the method?

     3kx +2y= 2xy

    Can you get x on its own in that equation?
    I don't get it
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    (Original post by knoxode)
    I don't get it
    So that equation I just gave you is the same type of question as the one we just worked through. If you can do that one on your own then it means you understand the method.

    The first thing to do if you want x on its own is to get all the x's on the same side of the equation. Can you do that?
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    (Original post by sindyscape62)
    So that equation I just gave you is the same type of question as the one we just worked through. If you can do that one on your own then it means you understand the method.

    The first thing to do if you want x on its own is to get all the x's on the same side of the equation. Can you do that?
    Nope.

    All I have is

    (3k + 2y)/2x = 2y
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    (Original post by knoxode)
    Nope.

    All I have is

    (3k + 2y)/2x = 2y
    Don't go too far ahead, just try the first step. There are two terms including x in that equation. What are they and how can you use adding or subtracting to get them on the same side?
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    (Original post by sindyscape62)
    Don't go too far ahead, just try the first step. There are two terms including x in that equation. What are they and how can you use adding or subtracting to get them on the same side?
    The two terms including x are:

    3kx and 2xy


    Therefore:

    3kx + 2y = 2xy => 2y = 2xy - 3kx

    I guess I could factorise to seperate x:

    2y = (2y - 3k)x

    Then divide through by (2y - 3k)

    2y/(2y-3k) = x
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    (Original post by knoxode)
    The two terms including x are:

    3kx and 2xy


    Therefore:

    3kx + 2y = 2xy => 2y = 2xy - 3kx

    I guess I could factorise to seperate x:

    2y = (2y - 3k)x

    Then divide through by (2y - 3k)

    2y/(2y-3k) = x
    Perfect!

    Maybe try a couple more to make sure you've got the hang of it.

     5x+7r-3=2xr

     2kx+4k=x - 2k

    Rearrange both to isolate x.
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    (Original post by sindyscape62)
    Perfect!

    Maybe try a couple more to make sure you've got the hang of it.

     5x+7r-3=2xr

     2kx+4k=x - 2k

    Rearrange both to isolate x.
    I have had a go at about another 15 questions.

    One that is confusing is

    Solve for u, (1/u) + (1/v) = (1/f)

    I have made it such that:

    fv + fu = vu

    and from there I have gotten to two different answers:

    u = fv/(v-f) and u = - (fv/(f-v))

    Are they both correct, or just one?
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    (Original post by knoxode)
    I have had a go at about another 15 questions.

    One that is confusing is

    Solve for u, (1/u) + (1/v) = (1/f)

    I have made it such that:

    fv + fu = vu

    and from there I have gotten to two different answers:

    u = fv/(v-f) and u = - (fv/(f-v))

    Are they both correct, or just one?
    Both of those answers are actually the same. If you take the first and multiply the top and bottom of the fraction by -1 then you get the second answer. It's the same as how
     \frac{-1}{-2} = \frac{1}{2}

    The first one is a bit neater though, so you'd usually give it in that form.
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    (Original post by sindyscape62)
    Both of those answers are actually the same. If you take the first and multiply the top and bottom of the fraction by -1 then you get the second answer. It's the same as how
     \frac{-1}{-2} = \frac{1}{2}

    The first one is a bit neater though, so you'd usually give it in that form.
    How do I convert between the two answers?
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    (Original post by knoxode)
    How do I convert between the two answers?
     u=\frac{fv}{v-f}

     u=\frac{fv}{v-f} \times \frac{-1}{-1}

     u=\frac{-fv}{-(v-f)}

     u=\frac{-fv}{-v--f}

     u=\frac{-fv}{f-v}

    Remember that multiplying the top and the bottom of a fraction by the same thing doesn't change, so I'm allowed to do that on line 2. Also, two minuses make a plus, as I used going from line 4 to 5.
 
 
 
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