ALG 19

Week 19: Having bitten the bullet in Week 18, we continue working with algebraic expression in this week's tasks. The focus again is on evaluating expressions, though we do also compare how their values change and we introduce two informal ways of determining when two expressions are equal.
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Monday: As with the expressions in Week 18, you might have to remind pupils of the conventions we use in writing expression, for example that 10e+3 means (10×e)+3. However, it would be worth first finding out what meaning pupils give to the expressions.
Before embarking on parts a), b) and c), it could be interesting to ask pupils, "Which expression is the larger?" It is quite likely that some pupils with think that the first expression is always larger (because of the 10 and the 2), while others think the second is always larger (because of the 3 and the 47). It might come as a surprise that the relative size of the expressions changes.
It is not expected that pupils will (or should) solve part c) using a formal method (such as taking 2e from both expressions). Rather, we expect pupils to use some form of trial and improvement, although this is quite challenging both because both expressions change on each trial and because the solution is not a whole number. For some pupils, this process will be fairly hit and miss, while others might get a sense that a change in e has a greater effect on the first expression than on the second (Why?). However, we anticipate that both sets of pupils will gain satisfaction from (eventually?) finding the solution!

When e = 2, the value of the first expression is considerably less than the value of the second (23 compared to 51). However, as e increases, the first expression rapidly catches up.... The two expressions have the same value when e = 5½.
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Tuesday: Here we compare our two expressions in a systematic way. Some pupils might already have done this, so this is an opportunity for drawing the class together.

Starting at e = 2, and increasing e by 1 each time, we see that the first expression rapidly approaches the value of the second and overtakes somewhere between e = 5 and e = 6.
Some pupils might notice that in the table, the difference in the value of the two expressions decreases by 8 each time (the differences are 28, 20, 12, 4, -4). Why is this happening?
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Wednesday: Here we compare two new expressions. We again use a systematic table of values and this time explicitly focus on the difference between the values and how this changes.

As u increases by 1, the difference in the value of the expressions decreased by 5 each time. When u = 3, the difference is 35, so it will take seven more steps of 5 to reduce this to 0. So this happens when u = 3+7 = 10.
This is an elegant (and quite powerful) method which might inspire some pupils. However, for others it might be taking things too far, so don't push this too hard. The solution u = 10 is relatively easy to spot using trial and improvement, and if some pupils are content with this, that's fine at this stage!
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Thursday: Here we introduce another method for equating the two expressions that involves matching elements in the expressions.

We are adding 3u+1 to 5u in one expression and to 50 in the other. So the two expresions will have the same value when 5u is equal to 50, ie when u = 10.
[Note: We are using the same expressions here as in Wednesday's task, so we already know the solution. It is worth making a virtue of this: we are not so much interested in the solution per se, but in gaining insight into methods (and into the nature of expressions) that we can make use of elsewhere.]
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Friday: These expressions may look daunting.... at first.

We have contrived to have two expressions that appear challenging but where the matching method used in Thursday's task might suddenly click for pupils, leading to a relatively simple solution.
Thus we can write the expressions as 23m + 7m + 33 and 7m + 33 + 2300, so they will have the same value when 23m and 2300 are equal, ie when m = 100.