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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
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    (Original post by vector12)
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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
    Start by factorising.
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    (Original post by vector12)
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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
    Your method is fine (if a bit unusual). Just take reciprocals of each side.
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    (Original post by vector12)
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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
    What do you notice about the RHS? Can you factorise it?

    Edit: As others have answered I will not post again!
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    You need to factorise to get: A = t(π + 5)
    Then divide A by (π + 5) to get A/(π + 5) = t
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    (Original post by vector12)
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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
    so basically factorise a t out on the right hand and the remainder you have in brackets you can then divide to get the red writing. SEE BELOW .
    Attached Images
     
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    As the others say - can you see a way to get just one t on the RHS?
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    Hope that helps mate
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    (Original post by vector12)
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    I divide both sides by t first, which gives me A/t = π +5

    However, dividing by A then gives me: 1/t = (π + 5)/A

    How do I then get the answer (in red)? And why can you do that?
    May be wrong but:
    1) factorise to A = t(π+5)
    2) divide by the whole bracket on both sides so left with A/(π+5) = t
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    (Original post by Mathematical_A)
    Your method is fine (if a bit unusual). Just take reciprocals of each side.
    Okay, so I have 1/t = (π + 5)/A

    How do I take reciprocals of both sides? And why are you able to do that?
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    (Original post by vector12)
    Okay, so I have 1/t = (π + 5)/A

    How do I take reciprocals of both sides? And why are you able to do that?
    Taking reciprocals means flipping the numerators and denominators of fractions on both sides of an equation.

    e.g. \displaystyle \frac{2}{3}=\frac{40}{60}

    and \displaystyle \frac{3}{2}=\frac{60}{40}
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    (Original post by Throwaway740)
    You need to factorise to get: A = t(π + 5)
    Then divide A by (π + 5) to get A/(π + 5) = t
    Thanks. How did you know to factorise it?
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    (Original post by Mathematical_A)
    Taking reciprocals means flipping the numerators and denominators of fractions on both sides of an equation.

    e.g. \displaystyle \frac{2}{3}=\frac{40}{60}

    and \displaystyle \frac{3}{2}=\frac{60}{40}
    Oh okay, thank you! I get it now, then you have t/1 on the LHS (which is obviously t).

    So when you flip the numerators and deniminators of both sides, does that only work if you have fractions on both sides? And if you had something without a fraction on one side, could you just put that side with 1 as a denominator, and then flip the fractions?

    eg. If I made up this situation, is this how it would be (if I worked backwards sort of thing)? Am I able to do this?
    Name:  IMG_1420.jpg
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    And if so, could I then say that 1 = hb/(a+c)? (if I wanted it in that form)

    NB: To clarify, thats a b and not a 6.
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    (Original post by vector12)
    Oh okay, thank you! I get it now, then you have t/1 on the LHS (which is obviously t).

    So when you flip the numerators and deniminators of both sides, does that only work if you have fractions on both sides? And if you had something without a fraction on one side, could you just put that side with 1 as a denominator, and then flip the fractions?

    eg. If I made up this situation, is this how it would be (if I worked backwards sort of thing)? Am I able to do this?
    Name:  IMG_1420.jpg
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    And if so, could I then say that 1 = hb/(a+c)? (if I wanted it in that form)
    Yes to all of those questions.
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    (Original post by Mathematical_A)
    Yes to all of those questions.
    Brilliant, thank you! I suppose taking a reciprocol just means making both sides the numerators and putting 1 as the denominator, right? And you can do that in any situation, right?
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    (Original post by vector12)
    Brilliant, thank you! I suppose taking a reciprocol just means making both sides the numerators and putting 1 as the denominator, right? And you can do that in any situation, right?
    That's not really the definition of a reciprocal.

    If x and y are reciprocals, then their product (the number you get when you multiply them) is 1.
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    (Original post by Mathematical_A)
    That's not really the definition of a reciprocal.

    If x and y are reciprocals, then their product (the number you get when you multiply them) is 1.
    Ok, but if I were to take a reciprocol, you are essentially putting 1/each side on each side aren't you?
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    (Original post by vector12)
    Ok, but if I were to take a reciprocol, you are essentially putting 1/each side on each side aren't you?
    Yes, you essentially put both sides as denominators and the numerator as 1.

    Then obviously you can simplify, but we skip the former step.
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    (Original post by RDKGames)
    Yes, you essentially put both sides as denominators and the numerator as 1.

    Then obviously you can simplify, but we skip the former step.
    Got it. Thanks a lot! Have a good evening!
 
 
 
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