There are a few questions I couldn't figure out, perhaps you can help me out.
1) Here is a number machine:
Input > multiply by 4 > subtract 8 > Output
When the input is a the output is b.
When the input is b the output is c.
Show clearly that c = 8(2a  5)
I don't understand where the 5 comes from. I thought it would be 40, since c = 16a  40.
2) I'm rationalising the denominator and simplifying:
77 divided by the square root of 11.
I multiplied it by square root of 11, which creates:
(77 x square root of 11) divided by 11.
Doesn't look simplified to me though, can't figure what I'm supposed to do next.
3) Solve the simultaneous equations:
y = 3x + 1
y squared = 4xsquared (as in 4 times x times x)  x + 7
I know y + ysquared = 2x + 8 + 4xsquared but that's as far as I could go. I don't know how this one's worked out.
4) Factorise 3nsquared (as in 3 x n x n) + 7n + 4
I think I'm in the ballpark with (3n + 7 + 4)(n+1) = 3 x n x n + 3n + 7n + 7 + 4n + 4
Few Higher Qs: algebra, denominator rationalising, simultaneous equation, factorising watch
 1
 2

ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 1
 05012013 16:30

Revision help in partnership with Birmingham City University

TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 2
 05012013 16:32
(1) because you have taken out 8 as a common factor

TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 3
 05012013 16:33
(2) cancel the 77 and the 11

TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 4
 05012013 16:35
(3) square the 3x+1 to give y^2
then put it equal to the expression that you have for y^2 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 5
 05012013 16:37
(4) not sure which ballpark you are in but that is just a straightforward factorise into 2 brackets question

claret_n_blue
 Follow
 0 followers
 16 badges
 Send a private message to claret_n_blue
Offline16ReputationRep: Follow
 6
 05012013 16:38
(Original post by ozzyoscy)
I don't understand where the 5 comes from. I thought it would be 40, since c = 16a  40.
(Original post by ozzyoscy)
(77 x square root of 11) divided by 11.
Doesn't look simplified to me though, can't figure what I'm supposed to do next.
(Original post by ozzyoscy)
4)
I think I'm in the ballpark with (3n + 7 + 4)(n+1) = 3 x n x n + 3n + 7n + 7 + 4n + 4
Mistake in factorising this. Have you learnt about the quadratic formula?
3) I can't do off the top of my head.
EDIT: Damn, beaten to it!
Ok for 3 then, like TenOfThem said. 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 7
 05012013 17:13
(Original post by claret_n_blue)
Look at what's outside the brackets.
(Original post by TenOfThem)
(2) cancel the 77 and the 11(Original post by claret_n_blue)
77 = 7 * 11
(Original post by TenOfThem)
(3) square the 3x+1 to give y^2
then put it equal to the expression that you have for y^2
(Original post by TenOfThem)
(4) not sure which ballpark you are in but that is just a straightforward factorise into 2 brackets question(Original post by claret_n_blue)
.
Mistake in factorising this. Have you learnt about the quadratic formula?
Spank you both so far.Last edited by ozzyoscy; 05012013 at 17:16. 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 8
 05012013 17:16
(Original post by ozzyoscy)
Oh of course. So y^2 = 5x^2 + 6  x, and y is the square root of that. Does it go further?
y^2 = (3x+1)^2
y^2 = 4x^2  x + 7
therefore
(3x+1)^2 = 4x^2  x + 7
expand, simplify, solve 
claret_n_blue
 Follow
 0 followers
 16 badges
 Send a private message to claret_n_blue
Offline16ReputationRep: Follow
 9
 05012013 17:25
(Original post by ozzyoscy)
Been doing mostly foundation stuff unfortunately, so it's iffy at the moment. I know how to multiply out brackets, tougher to do it the other way around. I'm getting the hang of where there are two operations, but this one has 3. I guess I have to put a minus somewhere, but knowing that there's just two brackets helps to know as a jumping off point.
Spank you both so far.
First you needed two numbers that multiply to give you your first number. The only option is 3 * 1 = 3. So right now your brackets are looking like this:
(3x + a)(x + b)
You know need to decided what two numbers multiply to give 4 and add to give 7. Obviously there are none, but don't forget that when expanding these brackets, you will multiply one of these numbers by 3. So now you are looking for two numbers that multiply to give for and two numbers where 3a + b = 7.
Can you think what these numbers are? The easiest way to do it would be to think of the two that multiply to give for and then trial and error from there. 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 10
 05012013 17:48
How about the (77 x square root of 11) divided by 11 a.k.a. 7 x (square root of 11 divided by 11)? What am I missing there?
(Original post by TenOfThem)
Not correct
y^2 = (3x+1)^2
y^2 = 4x^2  x + 7
therefore
(3x+1)^2 = 4x^2  x + 7
expand, simplify, solve
9x^2 = 4x^2  x + 6
Take 4x^2 from each side makes:
5x^2 = x + 6
That's as far as I got, and it doesn't look right.
(Original post by claret_n_blue)
The basic "method" when factorising a quadratic is looking for two numbers which multiply to give the last number (in your case its 4) and add to give the middle number (in your case its 7). Also you won't two other numbers that multiply to give you your first number. It's therefore easier if the first number is 1 and not 3 (like you have).
First you needed two numbers that multiply to give you your first number. The only option is 3 * 1 = 3. So right now your brackets are looking like this:
(3x + a)(x + b)
You know need to decided what two numbers multiply to give 4 and add to give 7. Obviously there are none, but don't forget that when expanding these brackets, you will multiply one of these numbers by 3. So now you are looking for two numbers that multiply to give for and two numbers where 3a + b = 7.
Can you think what these numbers are? The easiest way to do it would be to think of the two that multiply to give for and then trial and error from there.
From what you've said I've found it's (3n + 4)(n + 1)
I was in the ballpark! I just had to remove the 7 I put in the first bracket. 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 11
 05012013 17:54
(Original post by ozzyoscy)
How about the (77 x square root of 11) divided by 11 a.k.a. 7 x (square root of 11 divided by 11)? What am I missing there?
I got (3x+1)^2 which makes 9x^2 + 1.

ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 12
 05012013 18:02
No it does not = that
x x x = x^2
1 x 1 = 1
Thus (3x + 1)^2 = 9(x^2) +1, right? 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 13
 05012013 18:04
(3x+1)^2 = (3x+1)(3x+1) 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 14
 05012013 18:26
3x x 3x = 9x^2
3x x 1 = 3x
1 x 3x = 3x
1 x 1 = 1
(3x+1)^2 = 9x^2 + 6x + 1
So now I have:
y^2 = 9x^2 + 6x + 1
y^2 = 4x^2  x + 7
Then I remove like terms from both sides, and I have y^2 = 5x^2 + 6 + 5x 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 15
 05012013 18:28
(Original post by ozzyoscy)
Gah, maths is funny sometimes.
3x x 3x = 9x^2
3x x 1 = 3x
1 x 3x = 3x
1 x 1 = 1
(3x+1)^2 = 9x^2 + 6x + 1
So now I have:
y^2 = 9x^2 + 6x + 1
y^2 = 4x^2  x + 7
Then I remove like terms from both sides, and I have y^2 = 5x^2 + 6 + 5x
This makes no sense, I cannot see what you are doing
y^2=y^2 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 16
 05012013 19:22
9x^2  4x^2 = 5x^2, 6x  x = 5x, 7  1 = 6
So y^2 = 5x^2 + 5x + 6 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 17
 05012013 19:24
(Original post by ozzyoscy)
y^2 = 9x^2 + 6x + 1 = 4x^2  x + 7
Which you need to solve to find x
9x^2  4x^2 = 5x^2, 6x  x = 5x, 7  1 = 6
So y^2 = 5x^2 + 5x + 6 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 18
 05012013 19:40
(Original post by TenOfThem)
This is correct so you have
Which you need to solve to find x
This is nonsense 
TenOfThem
 Follow
 85 followers
 16 badges
 Send a private message to TenOfThem
Offline16ReputationRep: Follow
 19
 05012013 19:42
(Original post by ozzyoscy)
Yikes, no need to be a **** about it. If removing like terms isn't one of the processes to finding the answer, then I don't know the answer.
I am sure someone else will be along to help 
ozzyoscy
 Follow
 18 followers
 21 badges
 Send a private message to ozzyoscy
 Thread Starter
Offline21ReputationRep: Follow
 20
 05012013 19:48
This might help you out in the mean time.
 1
 2
Related discussions
 Edexcel C1 13th May 2015 *official thread*
 Year 12 Maths Help Thread
 GCSE Edexcel Maths 9th & 13th June
 Edexcel AS Mathematics C1/C2  18th/25th May 2016  Official ...
 Official Edexcel C1 & C2 June 2014 Thread
 Edexcel FP1 Thread  20th May, 2016
 Will the real TeeEm please stand up!
 STEP Prep Thread 2015
 STEP Prep Thread 2014
 STEP Prep Thread 2016
Related university courses

Mathematics with Statistics
University of Surrey

Economics, Statistics and Mathematics
Queen Mary University of London

Accounting, Finance and Mathematics
Lancaster University

Applied Mathematics
University of Aberdeen

Mathematics
University of Aberdeen

Mathematics
University of Winchester

Mathematics with Computing
Middlesex University

Mathematics and Statistics
University of Oxford

Mathematics and Statistics
University of Oxford

Mathematics for Finance (with a year abroad)
Swansea University
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.
 SherlockHolmes
 Notnek
 charco
 Mr M
 Changing Skies
 F1's Finest
 rayquaza17
 RDKGames
 davros
 Gingerbread101
 Kvothe the Arcane
 TeeEff
 The Empire Odyssey
 Protostar
 TheConfusedMedic
 nisha.sri
 claireestelle
 Doonesbury
 furryface12
 Amefish
 harryleavey
 Lemur14
 brainzistheword
 Rexar
 Sonechka
 TheAnxiousSloth
 EstelOfTheEyrie
 CoffeeAndPolitics
 an_atheist
 Labrador99
 EmilySarah00
 thekidwhogames
 entertainmyfaith
 Eimmanuel