P: 246

Hey guys,
Thank you for your suggestions but this is what I did in the lesson.
I started off by just laying some basic ground levels. i.e. I wrote x + 3 = 4 what is x, they all responded 1, so i moved on to x  3 = 4 etc.
They all said 'This is easy'. I said okay, why don't I give you a problem to do? They all said  is it algebra and I said 'no'.
The problem was thus, (stolen from Nrich!!)
George and Jim go to a sweet shop and want to buy a chocolate bar.
George needs 2 more p, Jim needs 50 more p. (p is pence  a unit of currency in UK)
They put their money together and still do not have enough.
How much is the chocolate bar.
They stared at me as though I was an alien. So, I started going through it and asked what we new. They instantly said that George needed 2p, and Jim needed 50p and that together they did not have enough.
I now said  okay I now start the clock on solving this problem. How shall we do it? Somebody said, try different amounts and get to an answer by changing it. I said fine  so we developed a table with Chocolate bar cost on the left, with the amount of money george must have in the next column ,and the amount of money jim had in the next, and in the final column, the addition of George and Jim.
Then I said  what is an approximate cost of a chocolate bar  somebody shouted out 30p so we used that. Obviously, it made Jim have 20p, I said, is this a problem and they instantly said it was as Jim cannot have 20p. So I said, what makes him have 20p when the chocolate bar costs 30p, they said it was the idea that Jim has 50p less.
We then reasoned that the smallest amount of money the chocolate bar could be was 50p, and so we used 50p , then 51p, and then 52p. For the first two we got numbers that fitted our conditions, but for 52p we didn't. So I asked, are 50p and 51p the only solutions  they said yes, probably. I said, were they sure and they said no.
I then said does anybody now a way to give an answer that must be write and that gives ALL the answers? 'Algebra' somebody whispered, and so I dived straight in!
At this point I was hoping someone would make the leap to an equation but nobody did, so I suggested, looking back a the 'x + 3' stuff, is there anyway we could use algebra here to show the link between the amount of money george has and the cost of a chocolate bar.
One child tentatively said, 'The chocolate bar is equal to george's money plus 2p' Some other members of the class looked a bit worried and said  thats not very 'mathy'. So I wrote it on the board and said what is an easier way to write the chocolate bar, they soon twigged I wanted a symbol so they chose C, I did the same for george's money and then Jim.
So we had.
c = g + 2
c= j + 50
I said good and what was the only piece of information we hadn't used? THey said we haven't used the adding together bit. So I asked them how would use it?
We got c = g + j
I said 'hmmmm, what would that mean' They soon saw the error but couldn't make the leap to a > sign. So said is C bigger than g + j....yes....and what is a mathematical way for writing bigger than? '>'
So we had
c = g +2
c = j + 50
c > g + j
At this point they had no idea where to go  and I was happy with that.
So I said, look at your example, x + 3 = 4, how many bits of that equation are not numbers? I then led them to the conclusion that you can only have 1 variable in an equation to solve it. So I said in c > g + j, which bit shall we leave? C was decided upon due to the fact it was in the other 2 equations.
So I said, we have c = g +2, how do we get g on its own (make the subject) so we can use what is left in the 3rd equation. Blank faces  they had never really rearranged equations. So i talked through that and soon we had
g = c 2
and j = c 50
Now, they were quite sheepish to add these into the inequality as they had never used an inequality for manipulation before but I told them that the rules for a > sign were the same as that for =. So we soon reached
c > 2c  52
This was manipulated to
52>c
So I said what does this tell us?
It means that the chocolate bar costs less than 52p.
'Good' I said. 'Now, looking at the other method we used, and used the idea that Jim has to have at least 0p, what is the minimum c can be?'
50p
so, what are the possible solutions? 50, 51.
The trial and improvement took 10 minutes and the algebra 15. So I said, shall we try again with different numbers?
And so I took another example through on the board in both methods.
Then i set them an example with these conditions  and timed them.
c = g + 4
and c = j + 6
This gives 15 solutions, and I set the most able girl the task of doing it by trial and improvement and the rest by algebra as a later test of how fast algebra is.
THe slowest kid finished in 5 minutes, the rest in about 3/4. The trial and improvement took about about 7, there were no calculators. I pointed out that this was only the third time they were dong things like this and they were still slow as compared to how fast you can become, but that they should realise how fast algebra is  they all agreed.
I was just wrapping up the lesson (it lasted about 1 hour) and a girl at the back who had been quite quiet piped up:
'Ben, algebra is more than quick'
'What do you mean?' I replied
'Its always right, gives you all the answers AND is quick' she said  and this reinforced for me the message with the class and the girl was only of average ability and so the class left buzzing for doing algebra and seeing a huge decrease in the time it took to solve a problem that looked huge before they started. Later on in the day they actually asked for more algebra!!
I thoroughly enjoyed it and think the kids learnt a lot.
NS
