A Level Maths

[kirovman]

So it is a matter of looking at the first 5 = a + b equation and working out in your head what a and b might equal, then seeing how that fits in with the 13 = 2a +3b equation to see if you're right?

I would do it on paper...you'll learn better that way. Yes, get an equation for a=.... then sub a into the other equation to find b.

 

[Ikerous]

So it is a matter of looking at the first 5 = a + b equation and working out in your head what a and b might equal, then seeing how that fits in with the 13 = 2a +3b equation to see if you're right?

If you have 5 = a + b and 13 = 2a + 3b
Then youd say b = 5 - a, and plug that in for b in 13 = 2a + 3b
Getting 13 = 2a + 15 - 3a, then you'd just solve for a
-2 = -a, so a = 2
And if a = 2, then 5 = 2 + b, so b is 3

 

[Razor]

If you have 5 = a + b and 13 = 2a + 3b
Then youd say b = 5 - a, and plug that in for b in 13 = 2a + 3b
Getting 13 = 2a + 15 - 3a, then you'd just solve for a
-2 = -a, so a = 2
And if a = 2, then 5 = 2 + b, so b is 3

Ok, so after you have the 13 = 2a + 15 - 3a part.

You're saying that the 2a and the 3a part cancel each other out, so you end up with 2a - 3a = -a? Where would you go from that point to find out that a = 2?

 

[Axyon]

Ok, so after you have the 13 = 2a + 15 - 3a part.

You're saying that the 2a and the 3a part cancel each other out, so you end up with 2a - 3a = -a? Where would you go from that point to find out that a = 2?

From that, you get -a = -2, therefore a = 2. Simple.

 

[DEATH eVADER]

Just started my alevel maths course and i am having a few difficulties understanding konstants.

Now, it says Y = k/square root of Z, and it gives Y = 10 and Z = 4.

So it then says 10 = k/square root of 4, then says that from that, you can work out that k = 20, now how does it work out that k = 20?

You should count yourself lucky, half of the people I talk to don't even know what differentiation and intergration are, which is what I'm doing. Half of the time, even I don't know what it is.

 

[Razor]

From that, you get -a = -2, therefore a = 2. Simple.

13 = 2a + 15 - 3a
13 = 15 + -a?
Change it round to 15 - 13 = a = 2?

 

[Razor]

You should count yourself lucky, half of the people I talk to don't even know what differentiation and intergration are, which is what I'm doing. Half of the time, even I don't know what it is.

It has been 5 - 6 years since i have done anything close to this difficult when it comes to Maths, that was when i did my GCSEs. So it is going to take a few weeks just to get back into my frame of mind but i do tend to really go at it as hard as possible as i need to complete both A levels by the beginning of next year at the latest.

 

[Axyon]

13 = 2a + 15 - 3a
13 = 15 + -a?
Change it round to 15 - 13 = a = 2?

Yeah, that's the other way of doing it. Either's fine.

 

[Razor]

Yeah, that's the other way of doing it. Either's fine.

Which way would you do it?

 

[Axyon]

Which way would you do it?

Honestly? Your way, since it doesn't primarily involve negatives, which can get very hard to handle in more complicated equations.

 

[JellyWorld]

Honestly? Your way, since it doesn't primarily involve negatives, which can get very hard to handle in more complicated equations.

Negative's aren't necessarily harder, it's just more tedious, and you're more likely to make a careless mistake somewhere.

 

[Axyon]

Negative's aren't necessarily harder, it's just more tedious, and you're more likely to make a careless mistake somewhere.

I said hard to handle, which relates to your comment about making an error somewhere.

 

[Razor]

Oh, i understand now

So the full way of working it out would be:

5 = a + b
13 = 2a + 3b

Turn 5 = a + b round to b = 5 - a, which would make 13 = 2a + 15 - 3a, where 15 - 3a = b. That becomes 13 = 15 - a, because 2a - 3a = -a.
13 = 15 - a turns into 15 - 13 = a = 2.

Going back to the very first equation of 5 = a + b, add the value for a being 2 would be 5 = 2 + b, turn that round to 5 - 2 = b, and you get b = 3.

That is the full way of working it out, wonderful.

 

[Dan]

Originally Posted by Razor
Ok, another prolbem to do with Partial Variation

It says y = ax + bz, where a and b are constants.

Where x = 1 and z = 1 and y = 5; and when x = 2 and z =3 and y = 13.

We substitute in y = ax + bz
5 = a + b and 13 = 2a + 3b, to solve the problem as a = 2 and b = 3, now how did they go from those 2 equations to solving it?

From what it looks like, there only explaining to me half of the working and leaving the other half in the dark :s.

wow, A levels sounds pretty easy if that's the kind of stuff you're doing. What grade is this? Wait till you get into mechanics and you have to solve 15 equations for 14 unknowns. Then you just put it into a matrix and use Gaussian elimination. You could do it with your 2 equations as well:
a+b=5
2a+3b=13

we turn this into a matrix form

1 1 | 5
2 3 | 13

basically the first collumn is a, the 2nd column is b and the 3rd column is the equivalent numbers. And the first row is equation 1 and the second row is equation 2.

This is just a simplified form of [1 1;2 3]*[a;b]=[5;13] if you were to plug it into matlab.

Then we do some transformations on the first matrix:

(original matrix)
1 1 5
2 3 13

(2nd row minus 2*1st row)
1 1 5
0 1 3

(1st row minus 2nd row)
1 0 2
0 1 3

and voila. 1a+0b=2 and 0a+1b=3
or a=2 and b=3

I know it sounds a lot more tedius but it will save you kajudles of time if you have 4 or 5 equations or more. And if you have a ti-83+ you can just plug the first matrix into your calculator and then use the rref command to spit out the answer.

 

[kirovman]

wow, A levels sounds pretty easy if that's the kind of stuff you're doing. What grade is this? Wait till you get into mechanics and you have to solve 15 equations for 14 unknowns. Then you just put it into a matrix and use Gaussian elimination. You could do it with your 2 equations as well:
a+b=5
2a+3b=13

we turn this into a matrix form

1 1 | 5
2 3 | 13

basically the first collumn is a, the 2nd column is b and the 3rd column is the equivalent numbers. And the first row is equation 1 and the second row is equation 2.

This is just a simplified form of [1 1;2 3]*[a;b]=[5;13] if you were to plug it into matlab.

Then we do some transformations on the first matrix:

(original matrix)
1 1 5
2 3 13

(2nd row minus 2*1st row)
1 1 5
0 1 3

(1st row minus 2nd row)
1 0 2
0 1 3

and voila. 1a+0b=2 and 0a+1b=3
or a=2 and b=3

Woah, it's his first day, take it easy. I think he's just getting the hang of the basics before he moves onto all the advanced stuff. Matrices don't come up until Further Maths anyway. Matrices method is simpler IF you know what matrices are.

 

[Razor]

wow, A levels sounds pretty easy if that's the kind of stuff you're doing. What grade is this? Wait till you get into mechanics and you have to solve 15 equations for 14 unknowns. Then you just put it into a matrix and use Gaussian elimination. You could do it with your 2 equations as well:
a+b=5
2a+3b=13

we turn this into a matrix form

1 1 | 5
2 3 | 13

basically the first collumn is a, the 2nd column is b and the 3rd column is the equivalent numbers. And the first row is equation 1 and the second row is equation 2.

This is just a simplified form of [1 1;2 3]*[a;b]=[5;13] if you were to plug it into matlab.

Then we do some transformations on the first matrix:

(original matrix)
1 1 5
2 3 13

(2nd row minus 2*1st row)
1 1 5
0 1 3

(1st row minus 2nd row)
1 0 2
0 1 3

and voila. 1a+0b=2 and 0a+1b=3
or a=2 and b=3

I know it sounds a lot more tedius but it will save you kajudles of time if you have 4 or 5 equations or more. And if you have a ti-83+ you can just plug the first matrix into your calculator and then use the rref command to spit out the answer.

It is my first day of a level maths, first 3 pages actually. Got to sort out a calculator and also remember the things i've forgotten since high school.

 

[kirovman]

It is my first day of a level maths, first 3 pages actually. Got to sort out a calculator and also remember the things i've forgotten since high school.

How many hours a day are you planning on putting in if you expect to get your full A-level before the end of the year?

From what I remember it was quite a lot of work, but of course I didn't actually do the learning until exam time, and even then I had to balance with 2 or 3 other subjects.

 

[ComradeBadger]

Haha, US math lessons are so much easier than UK Mathematics. Ask any Maths Professor

Anyway: just wait till you get to P3 'the noughter'

WHeeeee it's next monday and I haven't touched a paper of it

 

[kirovman]

Haha, US math lessons are so much easier than UK Mathematics. Ask any Maths Professor

Anyway: just wait till you get to P3 'the noughter'

WHeeeee it's next monday and I haven't touched a paper of it

I remember P3. I went to a gig 2 days before it in Manchester. And I did the P1 resit on the same day. Got 66% on P3, but got an A overall anyway.
It's got integration by parts and stuff hasn't it? Can't remember too well.

P2 was the best pure maths module.

 

[ComradeBadger]

Yeah I ripped up P2

I'm awful at Maths in general.. I'll be happy with a E/D

I'm an arts person and I'm taking Maths Physics and History.. (I fked up my History coursework - god I hate myself for that)

 

[baron insig]

yeah binomial expansion

tho formula rearraging is childs play

i do As Level physics! 2nd hardest A level i decided not to do maths Tooo hard!

 

[Axyon]

Heh, Pure maths is now 'Core' maths in the 2005 syllabus. I have no idea what's been changed other than that, apart from the fact that I think they made it a bit easier (although you couldn't tell). Sigh.

 

[ComradeBadger]

Yeah they split P1, P2 and P3 into C1, C2, C3 and C4

 

[marksmanHL2 :)]

Heh, Pure maths is now 'Core' maths in the 2005 syllabus. I have no idea what's been changed other than that, apart from the fact that I think they made it a bit easier (although you couldn't tell). Sigh.

Well theres no applied module in A2 now, but I think theres more pure maths to make up for it...


I don't know for sure though. Me and badge discussed this before. couldnt really find anything that both of us didnt do....

 

[Razor]

How many hours a day are you planning on putting in if you expect to get your full A-level before the end of the year?

From what I remember it was quite a lot of work, but of course I didn't actually do the learning until exam time, and even then I had to balance with 2 or 3 other subjects.

A good 15 - 20 hours per course, it isn't imperative that i do it by January, but it would make life a little easier. If i don't, what i might end up doing is doing alevels and raf training at the same time, not sure how that would work out so i was hoping to do it before i start the raf.

 

[Razor]

Kirovman or any of the other Maths geniuses, are any of you on msn messenger or aol? I promise i won't overuse your immense brains, just pick at them once and a while and i can be very good for sorting out relationship problems .

 

[Razor]

Expansion of Brackets

(2x - 1)(3x + 2) = 6x2 + 4x - 3x - 2 = 6x2 + x - 2.

Now i know that 6x and 4x comes from multiplying 2x with 3x and 2, but where does the - 3x and - 2 bit come from?

 

[ComradeBadger]

The result is 6x^2 + 4x -3x -2

First multiply 2x by 3x (giving 6x squared), then multiply -1 by 3x (giving the -3x ) and then multiply 2x by 2 (giving 4x ) and then the -1 by 2, giving -2.

I'm on MSN, I believe I have you on my list - I'm not too bad at maths, I'm currently revising it all - I'd be happy to help!

 

[ericms]

Expansion of Brackets

(2x - 1)(3x + 2) = 6x2 + 4x - 3x - 2 = 6x2 + x - 2.

Now i know that 6x and 4x comes from multiplying 2x with 3x and 2, but where does the - 3x and - 2 bit come from?

FOIL method.

First multiply the 2x by 3x and 2 then the -1 by 3x and 2.

 

[Razor]

If the roots of the equation 3x^2 + px + q = 0 are A, B, form the equation whose roots are 1/A, 1/B?

There is no explanation to how to do the equation, can anyone help. I know the answer is AB^2 + A^2B = AB(B + A) = q/3 (-p/3) = -pq/9. Can anyone explain just what is going on?