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    (Original post by RDKGames)
    1) Yes. b^2=a \rightarrow b=\pm\sqrt{a} (where a is whatever you want) but I'm unsure what you mean by "get it onto the other side"? If you square root b, then you square root the other side too, you don't move anything across.

    2) - We have: \frac{1}{\frac{R_1+R_2}{R_1R_2}}
    - Multiply top and bottom by R_1R_2 which gives: \frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}
    - Things cancel on the denominator and we are left with: \frac{R_1R_2}{R_1+R_2}

    3) Technically, you can just leave your answer as you found it firstly, but the answer in your book multiplied the fraction's numerator and denominator by -1 because it get rids of the negative on then numerator thus making it look simpler in a way.

    You found it to be: \frac{-fv}{f-v}
    Multiplying top and bottom by -1 doesn't change the value because: \frac{-fv}{f-v} \cdot 1 = \frac{-fv}{f-v} \cdot \frac{-1}{-1} = \frac{(-1)(-fv)}{(-1)(f-v)} = \frac{fv}{v-f}

    The tricks for 2) and 3) are literally the same, I hope you can see how this is helpful in achieving the answers in your book. If you multiply the numerator and denominator by the same factor, the quantity is unchanged because that very factor would cancel out anyway, however rather than cancelling it out, you can multiply across with other terms on the numerator and denominator.
    Thanks for your reply again.

    1) I'm not sure about the technical terms to explain what I done, but I meant what you said; that because I done something to one side, I done it to the other as well.

    2) I get everything in this question now - but, I dont understand how to multiply the whole fraction by R1R2; can you walk me through this too, pleaseee, as I can't get the answer that you get

    3) This question makes soo much more sense now - so thanks for your help!

    By the way, are you a maths teacher by any chance?
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    (Original post by Ak786454)
    Thanks for your reply again.

    1) I'm not sure about the technical terms to explain what I done, but I meant what you said; that because I done something to one side, I done it to the other as well.

    2) I get everything in this question now - but, I dont understand how to multiply the whole fraction by R1R2; can you walk me through this too, pleaseee, as I can't get the answer that you get

    3) This question makes soo much more sense now - so thanks for your help!

    By the way, are you a maths teacher by any chance?
    I'm not a maths teacher, but I've done some maths tutoring before

    1) If you square root a quantity, such as b there, then the other side requires to be the plus or minus of whatever you square rooted as well. As for the side on which b is, it doesn't require this sign.

    2) It's the same idea as in 3) you just use R_1R_2 rather than -1

    \rightarrow \frac{1}{\frac{R_1+R_2}{R_1R_2}} \cdot 1

    =\frac{1}{\frac{R_1+R_2}{R_1R_2}  } \cdot \frac{R_1R_2}{R_1R_2}

    = \frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}

    Notice within the main denominator the bottom of the smaller fraction can now be cancelled out because \frac{a}{b} \cdot b = a

    = \frac{R_1R_2}{R_1+R_2}
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    (Original post by BasharAssad)
    since it's in the form of ax^2 + bx and not ax^2 +bx + c (i.e. no c term) simply take out the common factor 'x', so you are correct.
    Thank you!
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    (Original post by RDKGames)
    I'm not a maths teacher, but I've done some maths tutoring before

    1) If you square root a quantity, such as b there, then the other side requires to be the plus or minus of whatever you square rooted as well. As for the side on which b is, it doesn't require this sign.

    2) It's the same idea as in 3) you just use R_1R_2 rather than -1

    \rightarrow \frac{1}{\frac{R_1+R_2}{R_1R_2}} \cdot 1

    =\frac{1}{\frac{R_1+R_2}{R_1R_2}  } \cdot \frac{R_1R_2}{R_1R_2}

    = \frac{1\cdot R_1R_2}{\frac{R_1+R_2}{R_1R_2} \cdot R_1R_2}

    Notice within the main denominator the bottom of the smaller fraction can now be cancelled out because \frac{a}{b} \cdot b = a

    = \frac{R_1R_2}{R_1+R_2}
    That makes soo much more sense now! Thanks soo much
    I swear you make maths look sooo easy!!!
    So what have you got your degree in then?
    By the way, you wouldn't be based in Birmingham by any chance would you? As I wouldn't mind a maths tutor to be fair.
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    (Original post by Ak786454)
    That makes soo much more sense now! Thanks soo much
    I swear you make maths look sooo easy!!!
    So what have you got your degree in then?
    By the way, you wouldn't be based in Birmingham by any chance would you? As I wouldn't mind a maths tutor to be fair.
    Degree? Hah, mate I only just finished Y13 this summer, going to study Maths for a degree in September though! I'm just used to teaching maths the way my teacher taught it, really. And unfortunately I am quite a distance from Brimingham but you can always ask here on the forums and people will explain stuff to you one way or another!
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    (Original post by RDKGames)
    Degree? Hah, mate I only just finished Y13 this summer, going to study Maths for a degree in September though! I'm just used to teaching maths the way my teacher taught it, really. And unfortunately I am quite a distance from Brimingham but you can always ask here on the forums and people will explain stuff to you one way or another!
    Woww, I never realised! I would've finished my Y13 this year, if I didn't get delayed by a year due to a few problems

    So what universities are you thinking of going to then?
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    (Original post by Ak786454)
    Woww, I never realised! I would've finished my Y13 this year, if I didn't get delayed by a year due to a few problems

    So what universities are you thinking of going to then?
    I've applied to Loughborough (firm) and Leicester so I'm waiting on results day to see what happens with those.
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    (Original post by RDKGames)
    I've applied to Loughborough (firm) and Leicester so I'm waiting on results day to see what happens with those.
    Ahh, that's nice then! Leicester is a nice place - my cousin went to that university too.

    Good luck with your results then!
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    Helloo everyone!

    Please can someone explain where I'm going wrong in this answer for this question as I'm coming up with a different answer to the one in the answer book:

    Simplify the following expression:
    (3x+2)^2 / 6x x x^4 / 6x+4

    - I would firstly cancel things out in order to get:
    (3x+2) / 6 x x^3 / 2x+2

    - I would then times things together in order to get:
    3x^4 + 2x^3 / 12x + 12

    - I could then factorise it in order to try and get a close answer to the on in the answer book:
    x^3(3x+2) / 12(x+1)

    ANSWER ON ANSWER BOOK:
    x^3(3x+2) / 12
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    (Original post by Ak786454)
    Helloo everyone!

    Please can someone explain where I'm going wrong in this answer for this question as I'm coming up with a different answer to the one in the answer book:

    Simplify the following expression:
    (3x+2)^2 / 6x x x^4 / 6x+4

    - I would firstly cancel things out in order to get:
    (3x+2) / 6 x x^3 / 2x+2

    - I would then times things together in order to get:
    3x^4 + 2x^3 / 12x + 12

    - I could then factorise it in order to try and get a close answer to the on in the answer book:
    x^3(3x+2) / 12(x+1)

    ANSWER ON ANSWER BOOK:
    x^3(3x+2) / 12
    It's your cancelling out that's causing the problem.
     \frac{(3x+2)^2}{6x} can't be simplified any further as there's no common factor- the same with he other fraction.
    To simplify a fraction you must divide the top and bottom by the same thing. In your working with the first fraction you've divided the top by  3x+2 and the bottom by  x These are not the same so you've changed the fraction.
    With the second one you've divided the top by  x and I'm not sure what you've done to the bottom. With this question multiplying together should be your first step.
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    (Original post by sindyscape62)
    It's your cancelling out that's causing the problem.
     \frac{(3x+2)^2}{6x} can't be simplified any further as there's no common factor- the same with he other fraction.
    To simplify a fraction you must divide the top and bottom by the same thing. In your working with the first fraction you've divided the top by  3x+2 and the bottom by  x These are not the same so you've changed the fraction.
    With the second one you've divided the top by  x and I'm not sure what you've done to the bottom. With this question multiplying together should be your first step.
    Thanks for your reply!

    Well, what I had actually done was that because it had '6x+4' on one side, (on the bottom) of the times sign and a '3x+2' on the other side, (at the top) of the times sign, I cancelled these two out, and I carried on cancelling in this way since this is how the book had cancelled out in an example question earlier on. So why can't I do that with this?
    And also how do I get to the answer in that case? This is messing with my head soo much!!
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    (Original post by Ak786454)
    Thanks for your reply!

    Well, what I had actually done was that because it had '6x+4' on one side, (on the bottom) of the times sign and a '3x+2' on the other side, (at the top) of the times sign, I cancelled these two out, and I carried on cancelling in this way since this is how the book had cancelled out in an example question earlier on. So why can't I do that with this?
    And also how do I get to the answer in that case? This is messing with my head soo much!!
    Ah, ok, sorry. That makes much more sense! What you've done is made a mistake with taking out the  3x+2 factor.
    When you divide  6x+4 by  3x+2 you are left with just 2, not  2x+2
    So after cancelling out you have
     \frac{3x+2}{6} \times \frac{x^3}{2}
    Hope that helps!
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    (Original post by sindyscape62)
    Ah, ok, sorry. That makes much more sense! What you've done is made a mistake with taking out the  3x+2 factor.
    When you divide  6x+4 by  3x+2 you are left with just 2, not  2x+2
    So after cancelling out you have
     \frac{3x+2}{6} \times \frac{x^3}{2}
    Hope that helps!
    That does make a bit more sense. So does that mean I'm supposed to divide 6x+4 by 3x+2 as a whole, if you see what I mean instead of dividing 6x by 3x then 4 by 2 to come out with 2x+2.

    Maths is soo confusing I swear!! I hope I get a decent grade in it :/
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    (Original post by Ak786454)
    That does make a bit more sense. So does that mean I'm supposed to divide 6x+4 by 3x+2 as a whole, if you see what I mean instead of dividing 6x by 3x then 4 by 2 to come out with 2x+2.

    Maths is soo confusing I swear!! I hope I get a decent grade in it :/
    Yes, you divide the whole expressions as in general, (a+b)/(c+d) does not equal (a/c)+(b/d).
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    (Original post by Ak786454)
    That does make a bit more sense. So does that mean I'm supposed to divide 6x+4 by 3x+2 as a whole, if you see what I mean instead of dividing 6x by 3x then 4 by 2 to come out with 2x+2.

    Maths is soo confusing I swear!! I hope I get a decent grade in it :/
    Yes. You don't separate them. (6x+4) and (3x+2) are expressions representing NUMBERS just as much as (10) or (14) are numbers, you cannot take them apart like that when both represent numerator and denominator of a fraction.
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    (Original post by Ak786454)
    That does make a bit more sense. So does that mean I'm supposed to divide 6x+4 by 3x+2 as a whole, if you see what I mean instead of dividing 6x by 3x then 4 by 2 to come out with 2x+2.

    Maths is soo confusing I swear!! I hope I get a decent grade in it :/
    Maybe it'd help to use brackets when you're dividing like that, just to check you're doing it right?
    Rewrite the fraction as \frac{x^4}{2(3x+2)}
    Now you can see that removing the bracket bit will leave just 2.

    The way you did it before you've turned 6x+4 into  (2x+2)(3x+2)
    If you try expanding that you can see it doesn't equal 6x+4. If you're ever not sure about a division do a multiplication with your answer and you should get what you started with.
 
 
 
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