I have often used the illustrated rabbit and balance situation as a starting point for discussion with teachers on solving algebraic equations.

It is quite a ridiculous idea - where do you get rabbits that are all the same size, shape and weight? How likely are they to sit like that on a set of scales? and so on - but it has always proved a rich source for some interesting methods and solutions, even if not all of them are correct. The consistent error is to think that “r” stands for rabbits rather than the weight of a rabbit.

Most teachers, but not all, remember some algebra from their schooldays and talk about taking 2r from both sides and then taking 3 from both sides or putting the “r“s on one side and the numbers on the other. Others try substituting numbers randomly. No one, apart from my son some 16 years ago while doing some homework, has come up with the idea of adding to both sides.

He reasoned that if you bought two more rabbits and added them to the right-hand side and then got four 1kg weights and added them to the left-hand side, the scales would still balance. Therefore 2r must equal 4kg and 1r = 2kg.

You can use this context to solve other combinations of rabbits and weights, except that each term of the equation must always be positive.

However, faced with something like 3r - 9 = 39 - 5r, you cannot use the context of weights and rabbits, as it would make no sense at all.

You therefore need another context. THOANs - “thinking of a number” type problems - are an alternative. A reasoning for 3r - 9 = 39 - 5r could be: I am thinking of a number, I multiply it by 3 and subtract 9. I would get exactly the same answer if I had multiplied it by 5 and subtracted the result from 39. What was my number?

Adding rabbits or kilogram weights to either side of some balancing scales is practical; adding some amount of unknown numbers to one side of an equation makes no sense at all.

In order to find the solution you have to work with symbols in a purely mathematical, algebraic way. And that is not an easy process to teach or to learn.

I will come back to this equation and a method for solving it later on. In the meantime, here is a possible method for solving missing number problems.

I have never seen anyone teach it, and it is not well known that it is possible to solve missing number problems by using a maximum of four (or even three) “trial and improvement” substitutions.

Here is a missing number problem: 1794 V ? = 39. This is very simple to solve providing we know that dividing 1794 by 39 would give you the missing number you require.

It may not be immediately obvious that this is the particular calculation you have to do - we need a good understanding of maths to recognise this.

An alternative method is to try dividing 1794 by different numbers until the answer 39 appears. However, this could take a long time. What follows is a much quicker “trial and improvement” method.

The only information you have to go on is the two numbers 1794 and 39 and the one operational sign, in this case division.

There are a limited number of operations you can carry out with two numbers. You can add them, subtract them, multiply them or divide them, but because the operational sign is division, you can safely assume that the solution will not involve addition or subtraction. So, four possibilities are: 1794 V 39 or 39V 1794 or 39 x 1794 or (but not necessary, as multiplication is commutative) 1794 x 39.

The answers are:

1794 V 39 = 46

39 V 1794 = 0.021739 (correct to 6 decimal points)

39 x 1794 = 69966

One of the numbers in bold is the missing number required in the equation 1794 V ? = 39. We now have to decide which one.

Common sense says 46 is the number we require, but we could try substituting 46, then 0.021739 and then 69966 for the question mark to see which one of the divisions gives us 39.

Keeping this process in mind, let us now go back to the equation3r - 9 = 39 - 5r.

This can also be solved by using a maximum of two specific “trial and improvement” substitutions.

To begin with, this is what happens when we keep substituting r = 1, r = 2,r = 3 and so on into the equation.

LHS +ve difference RHS

between RHS

amp; LHS

Substitute r = 1 - 6 40 34

Substitute r = 2 - 3 32 29

Substitute r = 3 0 24 24

Substitute r = 4 3 16 19

Substitute r = 5 6 8 14

Substitute r = 6 9 0 9

We find that when r = 6 the LHS = RHS and therefore r = 6 is the solution required. However it is possible to find the solution using the first two substitutions only.

LHS +ve difference RHS

betweenRHS amp; LHS

Substitute r = 1 - 6 40 34

Substitute r = 2 - 3 32 29

Subtract the +ve differences: 40 - 32 = 8

This tells you that the differences will keep reducing by 8 each time.

Divide the first difference by this result: 40 V 8 = 5

This tells you there will be five more substitutions to reach zero.

Add 1 to this last answer: 5 + 1 = 6

This means there will be six substitutions to get the LHS to equal the RHS. Therefore, r = 6.

We could also apply the “adding” method used in the rabbits example to solve this particular equation. If you add 48 to the LHS and 8r to the RHS the equation will still be balanced. Therefore 8r = 48 and 1r = 6.

The above methods may not be common, or even an acceptable alternative to the traditional methods, but they are novel.

Peter Critchley is a former advisory teacher and numeracymanager in Suffolk