Maths Homework - My boys (not mine - honest!!) [modified]

[Luc Pattyn]

Hi Malcolm,

Malcolm Smart wrote:

My boys (not mine - honest!!)

Seems to me the difference is only a matter of time.

Malcolm Smart wrote:

b+1 = b.a2

almost there. It is hard to believe you get stuck at this point. Move all the terms containing b to one side, all other to the other and you get b.a2 - b = 1 :)

Luc Pattyn [Forum Guidelines] [My Articles]

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[Malcolm Smart]

Hi Luc! Had a sneeky feeling you might turn up.

Luc Pattyn wrote:

Seems to me the difference is only a matter of time.

:laugh: Can't believe how much I've forgotten (or never knew!)

Luc Pattyn wrote:

almost there. It is hard to believe you get stuck at this point. Move all the terms containing b to one side, all other to the other and you get b.a2 - b = 1

We've been here - we can't the get a2 - 1. You managed to tease an answer out me before regarding lemmings so I know you're not gonna give me the answer, but there is some obvious algebraic rule I am missing. I need to get down to a single b on one side of the equation. If I take the a2 over to get 1/a2, I am left on the left hand side with b-b, which equals zero, so that seems wrong. We took the a2-b over, giving us b= 1/a2-b, which is very nearly there, except it should ne a2-1, not a2-b. (maths isn't my main thing - you can tell!) My son is convinced the book is wrong - can you at least confirm b = 1 / (a2 - 1) is correct?

so you answer don't be scared of failure The only failure is never to try Things You've Never Done - Passenger -2008

[Luc Pattyn]

I'll never forget those lemmings that approach a lethal ravine at high speed, suddenly fall down strictly vertically, at a constant speed, and survive it all.

Malcolm Smart wrote:

b = 1 / (a2 - 1)

is correct. Fact: you can't "take a2-b over", taking over basically means you add something both left and right in such a way that it cancels out something on one side. Now what would you add to get rid of b.a2? certainly not a2! Hint: if x.y = z how much is x? so try bringing your equation in a similar form. :)

Luc Pattyn [Forum Guidelines] [My Articles]

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This month's tips: - before you ask a question here, search CodeProject, then Google; - the quality and detail of your question reflects on the effectiveness of the help you are likely to get; - use PRE tags to preserve formatting when showing multi-line code snippets.

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[Malcolm Smart]

I either get b-b on the left, or b back on both sides! (a2 = a2) <-- saves typing html tags for sup

 b.a2-b=1
(b.a2)-b=1 
b.a2 =  1 + b
b = (1 + b) / a2

but don't know how this then gets tob = 1 / (a2 -1)
There must be something so obvious I am missing, but I have to give up at this point (my son already has - he will learn soon that isn't an option sometimes in the real world) - my own work is pressing as I am having to work weekends at the moment. I will get this solved, but have to leave it until in the week. Thanks Luc - appreciate your help.

so you answer don't be scared of failure The only failure is never to try Things You've Never Done - Passenger -2008

[Luc Pattyn]

You can factor the lefthand side b.a2-b=1 into b and something, then you have the form x.y=z which just requires a division. :)

Luc Pattyn [Forum Guidelines] [My Articles]

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This month's tips: - before you ask a question here, search CodeProject, then Google; - the quality and detail of your question reflects on the effectiveness of the help you are likely to get; - use PRE tags to preserve formatting when showing multi-line code snippets.

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[Malcolm Smart]

Luc Pattyn wrote:

You can factor the lefthand side b.a2-b=1 into b and something

Waheeeyyy!!! I just called him down and asked him to factorise b.a2-bb(a2-1)
which givesb(a2-1) = 1 b = 1 / (a2 - 1)
It's all about applying what he knows. He didn't think to factorise ( I didn't know you could!). Thanks a million Luc. He only has another 12 to do!! I'm going to have to really get into this stuff - it's actually quite cool.

so you answer don't be scared of failure The only failure is never to try Things You've Never Done - Passenger -2008

[Luc Pattyn]

You're welcome.

Malcolm Smart wrote:

It's all about applying what he knows

Of course. Don't apply what you don't know, but don't forget to apply what you do know; what is the purpose of knowing something, if you don't apply it...

Malcolm Smart wrote:

I'm going to have to really get into this stuff

Knowledge is hereditary, it will find its way up or down. :-D

Luc Pattyn [Forum Guidelines] [My Articles]

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This month's tips: - before you ask a question here, search CodeProject, then Google; - the quality and detail of your question reflects on the effectiveness of the help you are likely to get; - use PRE tags to preserve formatting when showing multi-line code snippets.

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[kinar]

Of course this is probably way too late to help but... The key here is that you need to isolate b starting from your step: (b+1) / b = a^2 this is equal to: (b+1) * (1/b) = a^2 multiply out left side: (b/b) + (1/b) = a^2 reduce (b/b): 1 + (1/b) = a^2 subtract 1 from both sides: 1/b = a^2 - 1 multiply both sides by b: b/b = (a^2 - 1) * b multiply both sides by 1/(a^2 -1): (1/(a^2 - 1))*(b/b) = (a^2 - 1) * (1/(a^2 - 1)) * b reduce both sides: 1/(a^2 - 1) = b

[SilimSayo]

Simpler way (b+1)/b=a^2 (b/b) + (1/b)=a^2 1 + (1/b) =a^2 (1/b)= a^2 - 1 => b= 1/(a^2-1)

[kinar]

Thats the same thing as I posted. I just included all the steps that you left out which would be required by a math teacher who is actually teaching this stuff Although I could have sworn that the entire other thread didn't exist from 2 weeks before walking them through it when I posted this.... :(