Questions about algebraic transformation

Announcements
    • Thread Starter
    Offline

    1
    ReputationRep:
    Hi, I have a few problems understanding the following;

    1-n = x/y becomes n = 1-(x/y)

    I don't understand this, I would have expected this: -n=(x/y)-1

    It's the n that's negative not the 1, so why it would suddenly not be the case is confusing me.

    --------------------

    again, I have problems with this;

    1-(p/q) = (q-p)/q

    :mad:

    any help would be greatly appreciated, I've enrolled onto an HNC and this is a big jump up to the maths I remember doing.
    Offline

    2
    ReputationRep:
    (Original post by Brainfrozen)
    Hi, I have a few problems understanding the following;

    1-n = x/y becomes n = 1-(x/y)

    I don't understand this, I would have expected this: -n=(x/y)-1

    It's the n that's negative not the 1, so why it would suddenly not be the case is confusing me.
    If you take -n=(x/y)-1 and multiply both sides by -1 what do you end up with?

    again, I have problems with this;

    1-(p/q) = (q-p)/q
    In order to manipulate the LHS we need both parts to have the same denominator i.e. we want them both as fractions with q on the bottom. How can you rewrite 1 so it has a q on the bottom? (Or maybe look at it as, what fraction with q as the denominator/bottom cancels to 1)
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by Elzar)
    If you take -n=(x/y)-1 and multiply both sides by -1 what do you end up with?



    In order to manipulate the LHS we need both parts to have the same denominator i.e. we want them both as fractions with q on the bottom. How can you rewrite 1 so it has a q on the bottom? (Or maybe look at it as, what fraction with q as the denominator/bottom cancels to 1)
    n = -(x/y) +1

    I understand now

    (q/q) = 1

    If only my lecturers were as rapid at answering I wouldn't have so many problems!

    Expect more silly questions!
    • Thread Starter
    Offline

    1
    ReputationRep:
    Another problem with transformation, I've moved onto series and understand this ok, however its the transformation that's doing my head in;

    408=(n/2)(2×12+(n−1)×4)

    I just don't understand how I'm doing this so wrong each time, you think you get somewhere and hit a brick wall

    I'll show you how I have been trying it;

    408 x 2 = n(2×12+(n−1)×4)

    816/n = 24+(n−1)×4

    n = 816/(24+(n−1)×4)

    4n = 816/(24+n−1)

    (4n)n = 816/234

    n^2 = etc

    very wrong
    Offline

    3
    ReputationRep:
    (Original post by Brainfrozen)
    Another problem with transformation, I've moved onto series and understand this ok, however its the transformation that's doing my head in;

    408=(n/2)(2×12+(n−1)×4)

    I just don't understand how I'm doing this so wrong each time, you think you get somewhere and hit a brick wall

    I'll show you how I have been trying it;

    408 x 2 = n(2×12+(n−1)×4)

    816/n = 24+(n−1)×4

    n = 816/(24+(n−1)×4)

    4n = 816/(24+n−1)

    (4n)n = 816/234

    n^2 = etc

    very wrong
    You're correct up to the third line. The problem with the fourth is that you multiply both sides with 4 but the 24 is not affected.

    \displaystyle 24+4(n-1)= 4(\frac{24}{4})+4(n-1)=4(\frac{24}{4}+n-1)=4(6+n-1)=4(n+5)

    and this would be the denominator. And NOW you can multiply freely by 4 on both sides.

    Though of course, your method is way too long winded.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (Original post by RDKGames)
    You're correct up to the third line. The problem with the fourth is that you multiply both sides with 4 but the 24 is not affected.

    \displaystyle 24+4(n-1)= 4(\frac{24}{4})+4(n-1)=4(\frac{24}{4}+n-1)=4(6+n-1)=4(n+5)

    and this would be the denominator. And NOW you can multiply freely by 4 on both sides.

    Though of course, your method is way too long winded.
    Thanks for that RDK, but I'm not moving forward with this, I understand what you mean, but I'm struggling to understand how to isolate n when there's two n terms and one is squared. Factorising doesn't seem to help the issue.

    Could you advise me on the methodology to solve for n? I'd rather understand the thinking behind the method rather than see how to solve it if you know what I mean.
    Offline

    3
    ReputationRep:
    (Original post by Brainfrozen)
    Thanks for that RDK, but I'm not moving forward with this, I understand what you mean, but I'm struggling to understand how to isolate n when there's two n terms and one is squared. Factorising doesn't seem to help the issue.
    You can factorise it since you're solving it. Otherwise, complete the square and use the \pm sign to display the two solutions.
    • Thread Starter
    Offline

    1
    ReputationRep:
    I don't understand this at all I'm afraid.
    Offline

    3
    ReputationRep:
    (Original post by Brainfrozen)
    I don't understand this at all I'm afraid.
    You need to get a quadratic in the form an^2+bn+c=0 from what you are given, and solve it for n. You don't really make it clear on what you need to do with it (considering you mentioned transformations, unless you meant algebraic manipulations instead).

    408=\frac{n}{2}[2\cdot 12+4(n-1)]

    \Rightarrow 816=n[24+4n-4]

    \Rightarrow 816=n(4n+20)

    \Rightarrow 816=4n^2+20n

    \Rightarrow 0=4n^2+20n-816

    \Rightarrow n^2+5n-204=0

    Then complete the square on it, or factorise it.
    • Thread Starter
    Offline

    1
    ReputationRep:
    Don't know mate, I just need to solve for n, it should equal 12 as u know.

    I have gaps in my understanding, can understand medium difficulty but tend to get tripped up on silly things.

    Eg, polynomial division, once I went back and remembered how to do long division it was easy.
    Offline

    3
    ReputationRep:
    (Original post by Brainfrozen)
    Don't know mate, I just need to solve for n, it should equal 12 as u know.

    I have gaps in my understanding, can understand medium difficulty but tend to get tripped up on silly things.

    Eg, polynomial division, once I went back and remembered how to do long division it was easy.
    Well go ahead an remind yourself of how to solve quadratics then you'll have it.
    • Thread Starter
    Offline

    1
    ReputationRep:
    (n+17)(n-12)=0

    n must = -17 or 12

    Offline

    3
    ReputationRep:
    (Original post by Brainfrozen)
    (n+17)(n-12)=0

    n must = -17 or 12

    Yep. If you want it as n equals something (single equals sign) then from completing the square you would have \displaystyle n=\frac{-5\pm 29}{2}
    • Thread Starter
    Offline

    1
    ReputationRep:
    1/Z =(1/A) +(1/B)

    ∴ Z=(A*B)/(A+B)

    Can someone show me the steps required to get from the first formula to the last?
    Offline

    2
    ReputationRep:
    (Original post by Brainfrozen)
    1/Z =(1/A) +(1/B)

    ∴ Z=(A*B)/(A+B)

    Can someone show me the steps required to get from the first formula to the last?
    Add the fractions on the RHS

    1/Z=1/A+1/B = (A+B)/AB
    Take the reciprocal of both sides
    (Take 1 and divide it by both sides)
    Z=AB/(A+B)
 
 
 
Write a reply… Reply
Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. Oops, you need to agree to our Ts&Cs to register
  2. Slide to join now Processing…

Updated: September 22, 2016
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

Poll
Which is the best season?
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
Study resources

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.