Listen to Jenny and Graeme talking together about the problem. [ audio ]

Encounters with simultaneous equations can become over-familiar, routine experiences for students. This type of problem causes a "stop and think" moment, requires some problem-solving ingenuity, and leads into a consideration of redundancy of information. Students might make a start by substituting some arbitrary $x$, $y$ values to get a feel for the problem and to grasp that the five expressions don't generally take the same value.

This is in contrast to expressions that are identities, for example $2 (x+y) - 3(x-y )$ and $5y - x$, where the two expressions take the same value for any $x$, $y$ combination. This idea is worth some discussion.

Questions or prompts: For a start:

$2 x + 3 y - 20$ equals $5 x - 2 y + 38$ ...

Could you find an $x$ , $y$ pair that works for two, for three, or for four of the expressions but not for all of them ?

Further ideas:

Make up a similar problem of your own.

Or extending that : can you create a similar problem with an ödd one out"? That is, one expression which does not equal the other four, which are equal for some specific

x

y

pair.

The following interesting account was sent in by a class teacher working withYear 8s in Maths Club at St Albans High School for Girls

Becky worked as follows:

2x + 3y -20 = 4x + 5y -72

(-2x to each side)

3y -20 = 2x + 5y - 72

(+72 to each side)

3y + 52 = 2x + 5y

(-3y to each side)

52 = 2x + 2y

Then, 5x - 2y + 38 = x - 4y + 108

(-x from each side)

4x - 2y + 38 = -4y + 108

(+4y to each side)

4x + 2y + 38 = 108

( -38 to each side)

4x + 2y = 70

Becky looked at the difference between these two equations and deduced

2x = 18, so x = 9

Then using one of her equations, and substituting x = 9 she found y = 17.

Ele and Sarah reached the same conclusion but started out by looking at

all the possible pairs of expressions. Then they selected the following two

as easiest to work with:

From 2x + 3y - 20 = 4x + 5y - 72

They deduced 26 = x + y

From 2x + 3y - 20 = x - 4y + 108

They deduced 128 = x + 7y

From these two equations they deduced 6y = 128 - 26

6y = 102

y= 17

Then from x + y = 26 they found x = 9

All three girls were then challenged to decide how much of the information

they needed to use to solve the problem. Their conclusion was only 3

statements were needed; they could have used A = B and A = C to deduce the

answer where A,B,C label different expressions given.