Hey there! Sign in to join this conversationNew here? Join for free

Fractions are going to prevent me from getting a grade A! Watch

    • Thread Starter
    Offline

    0
    ReputationRep:
    So I've trawled through bucket loads of past papers both official and Solomon and I can attemp every question without a problem... BUT...

    If the questions involve manipulating fractions I can't seem to do it, or I attempt it and end up with the wrong solution. This is the case when I'm solving simultaneous equations, or finding the stationary points, completing the square etc...

    A basic example... I end up with a quadratic whose solution is \frac{1}{3} and say 2

    I have no trouble in substituting x=2 into the original equation, lets say x^3+2x^2+3x+4 BUT when substituting x=1/3 I get into a mess... This is how I would work it out:

    At x=1/3,

    \frac{1}{3}^3+2\frac{1}{3}^2+3\f  rac{1}{3}+4

    \frac{1}{27}+\frac{2}{9} ...

    I would then solve both sets of fractions and then continue to find y... So at this point y would equal:

    \frac{52+9}{213}=\frac{61}{213}

    I would then continue to find y ...

    \frac{61}{213}+3\frac{1}{3}+4 ...

    As you can see this is extremely long winded and there's a lot of room for error. Also I don't cancel down where appropriate because I can't.

    Is there a faster way to finding y when x is a fractional value?

    [latex]
    • PS Helper
    Offline

    14
    Well you can notice that the lcm of 3,9,27 is 27 (and even if it wasn't, you should have had 243 instead of 213). But this means that the common denominator you can use is 27.

    I'm slightly confused about your problem though; x^3+2x^2+3x^3+4 isn't a quadratic, and it doesn't have 2 or \dfrac{1}{3} as a root... so I'm slightly puzzled as to why you're substituting x=\dfrac{1}{3}.
    Offline

    2
    ReputationRep:
    Is that a real example or a made up example? Just curious, because usually there aren't fractions as bad as that the entire way through. Also, you shouldn't get 61/213 ... get the answer with a calculator and see where you've gone wrong (because your answers before the "..." are fine)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by nuodai)
    Well you can notice that the lcm of 3,9,27 is 27 (and even if it wasn't, you should have had 243 instead of 213). But this means that the common denominator you can use is 27.

    I'm slightly confused about your problem though; x^3+2x^2+3x^3+4 isn't a quadratic, and it doesn't have 2 or \dfrac{1}{3} as a root... so I'm slightly puzzled as to why you're substituting x=\dfrac{1}{3}.
    Sorry I forgot to mention that I obtained x=2 and x=1/3 when I differentiated the cubic and obtained a quadratic.

    (Original post by linkdapink)
    Is that a real example or a made up example? Just curious, because usually there aren't fractions as bad as that the entire way through. Also, you shouldn't get 61/213 ... get the answer with a calculator and see where you've gone wrong (because your answers before the "..." are fine)
    It's a C1 exam and actually these are very common throughout the paper tbh.
    Offline

    2
    ReputationRep:
    (Original post by Student#254)
    It's a C1 exam and actually these are very common throughout the paper tbh.
    Woah C1's got a LOT harder since I last did it (only about 2 years ago). Could you have made a mistake with the differentiation? Can you tell me where to find the question so I can have a look?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by linkdapink)
    Woah C1's got a LOT harder since I last did it (only about 2 years ago). Could you have made a mistake with the differentiation? Can you tell me where to find the question so I can have a look?
    Oh FFs the exampes aren't accurate :P

    Ok erm... Classic example here, look at question 8:

    http://www.ocr.org.uk/download/pp_08...721_answer.pdf
    Offline

    2
    ReputationRep:
    (Original post by Student#254)
    Oh FFs the exampes aren't accurate :P

    Ok erm... Classic example here, look at question 8:

    http://www.ocr.org.uk/download/pp_08...721_answer.pdf
    What nuodai has said is right.

    It's all about finding the LCM for the denominator. If you have 3 fractions, with denominators 27, 9 and 3 the LCM is 27. You don't need to be multiplying any numbers together for that. In terms of working it out; clear and concise solutions will always help. Be systematic in your working and do all the steps; that way you can clearly see what you've done and what you're trying to get to.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by dknt)
    What nuodai has said is right.

    It's all about finding the LCM for the denominator. If you have 3 fractions, with denominators 27, 9 and 3 the LCM is 27. You don't need to be multiplying any numbers together for that. In terms of working it out; clear and concise solutions will always help. Be systematic in your working and do all the steps; that way you can clearly see what you've done and what you're trying to get to.
    Ok Question 8:

    y=x^3+x^2-x+3

    \frac{dy}{dx}=3x^2+2x-1

    3x^2+2x-1=0

    (3x-1)(x+1)

    x=\frac{1}{3} and x=-1

    at x=\frac{1}{3}, y=\frac{1}{3}^3+\frac{1}{3}^2-\frac{1}{3}+3

    y=\frac{1}{27}+\frac{1}{3}+3

    How would I solve this to find y using a LCD of 27?

    If I get grasp using a LCD then I will be scoring nearly 100% on all C1 papers. At the moment I seem to be misunderstanding a basic fundamental concept
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by nuodai)
    Well you can notice that the lcm of 3,9,27 is 27 (and even if it wasn't, you should have had 243 instead of 213). But this means that the common denominator you can use is 27.

    I'm slightly confused about your problem though; x^3+2x^2+3x^3+4 isn't a quadratic, and it doesn't have 2 or \dfrac{1}{3} as a root... so I'm slightly puzzled as to why you're substituting x=\dfrac{1}{3}.
    The mark scheme says the answer is 76/27. I don't get that...

    y=\frac{1}{27}+\frac{1}{3}+3

    \frac{30}{27}+\frac{1}{3}= \frac{101}{27}

    \frac{101}{27}+{3}{1}...

    This does not get the right answer even though my method seems correct??
    Offline

    2
    ReputationRep:
    (Original post by Student#254)
    Ok Question 8:

    y=x^3+x^2-x+3

    \frac{dy}{dx}=3x^2+2x-1

    3x^2+2x-1=0

    (3x-1)(x+1)

    x=\frac{1}{3} and x=-1

    at x=\frac{1}{3}, y=\frac{1}{3}^3+\frac{1}{3}^2-\frac{1}{3}+3

    y=\frac{1}{27}+\frac{1}{3}+3

    How would I solve this to find y using a LCD of 27?

    If I get grasp using a LCD then I will be scoring nearly 100% on all C1 papers. At the moment I seem to be misunderstanding a basic fundamental concept
    Don't worry about it

    Let's start from here

    x=\frac{1}{3}, y=(\frac{1}{3})^3+(\frac{1}{3})^  2-\frac{1}{3}+3

    That then gives y= \frac{1}{27} + \frac{1}{9} -\frac{1}{3} +3

    From this we need some way to combine these fractions, which means all the denominators must be the same. We can see that 27, 9 and 3 have a common multiple of 27 i.e. 27(1)=27, 9(3)=27, 3(9)=27. And so now we apply this to the fractions and make everything over 27. So first we have \frac{1}{27} This is already over 27 and so we don't need to do anything. \frac{1}{9} Well, to get from 9 to 27 we multiply by 3; and whatever you do to the top you do to the bottom, so it becomes \frac{3}{27} And you can apply the same idea to the rest. Also, remember that 3=\frac{3}{1}. From there it's just adding and taking away fractions.

    Always try to find that lowest common multiple in the denominators. It makes life alot easier. Having said that, some examples may not have a 'low' LCM and it may be that you do just have to multiply all your numbers together to find one. If that happens just use the same ideas and work through it carefully and slowly.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by dknt)
    Don't worry about it

    Let's start from here

    x=\frac{1}{3}, y=(\frac{1}{3})^3+(\frac{1}{3})^  2-\frac{1}{3}+3

    That then gives y= \frac{1}{27} + \frac{1}{9} -\frac{1}{3} +3

    From this we need some way to combine these fractions, which means all the denominators must be the same. We can see that 27, 9 and 3 have a common multiple of 27 i.e. 27(1)=27, 9(3)=27, 3(9)=27. And so now we apply this to the fractions and make everything over 27. So first we have \frac{1}{27} This is already over 27 and so we don't need to do anything. \frac{1}{9} Well, to get from 9 to 27 we multiply by 3; and whatever you do to the top you do to the bottom, so it becomes \frac{3}{27} And you can apply the same idea to the rest. Also, remember that 3=\frac{3}{1}. From there it's just adding and taking away fractions.

    Always try to find that lowest common multiple in the denominators. It makes life alot easier. Having said that, some examples may not have a 'low' LCM and it may be that you do just have to multiply all your numbers together to find one. If that happens just use the same ideas and work through it carefully and slowly.
    I get:

    y=\frac{1}{27}+\frac{3}{27}+ \frac{9}{27}+ \frac{3}{1}

    y=\frac{13}{27}+\frac{3}{1}= \frac{81}{27}

    The mark scheme staes \frac{76}{27}

    Somehow my working is wrong?

    Your post really helped a lot though! :P
    Offline

    19
    ReputationRep:
    You're multiplying the denominators to get your new denominator every time. You only do that if there's no lower common denominator to use. Where possible, you should use a lowest common denominator to simplify things.

    1/3 + 1/9 = (3/9) + 1/9 = 4/9
    So you shouldn't be using 27 on your denominator here

    3 + 1/3 + 1/9 = 27/9 + 3/9 + 1/9 = 31/9
    Again, you shouldn't be using 27 on your denominator here

    1/5 + 1/7 = 7/35 + 5/35 = 12/35
    This is where you multiply 7 and 5 to get a common denominator; there is no lower common denominator to use

    Another good example:

    1/8 + 1/6 = 3/24 + 4/24 = 7/24
    I spot that 24 is the lowest common denominator; no need to use 8x6 = 48.
    Offline

    19
    ReputationRep:
    1/27 + 1/9 - 1/3 + 3
    = 1/27 + 3/27 - 9/27 + 81/27
    = 76/27

    The lowest common denominator of 27, 9, 3 and 1 is 27; so we use that.
    Offline

    19
    ReputationRep:
    (Original post by Student#254)
    y=\frac{13}{27}+\frac{3}{1}= \frac{81}{27}
    No!

    13/27 + 3/1 = 13/27 + 81/27 = 94/27 ... With your working
    • Thread Starter
    Offline

    0
    ReputationRep:
    OMG!!! This really helped a lot.

    I feel as if everything has just clicked! OMG... So awesome... Lol.

    Thank you lol.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Physics Enemy)
    No!

    13/27 + 3/1 = 13/27 + 81/27 = 94/27 ... With your working
    I know I made a sign error... added instead of minus :P
    Offline

    19
    ReputationRep:
    (Original post by Student#254)
    I know I made a sign error... added instead of minus :P
    Yes but even when you messed up your working, you added the fractions wrong at the end: 13/27 + 3 = 13/27 + 81/27 = 94/27. You wrote 81/27 as the answer; seems like you forgot to add on 13/27.

    The idea: we always want a lowest common denominator when adding/subtracting fractions. Sometimes we're forced into multiplying the denoms to make one. On other occassions, we just manipulate some of the fractions and then combine. Don't keep using the cross-multiply trick every time.

    If you cross multiply 1/27 and 1/3, then you've obviously missed the fact that 3 fits into 27 nicely (9 times). You should use that to your advantage and write 1/3 as 9/27.

    Just like 1/8 and 1/6 ... both 8 and 6 fit nicely into 24; use that to your advantage.
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.