Finding a Simultaneous Solution: Previously Known as Solving Simultaneous EquationsAuthor(s): Tom LeddySource: Mathematics in School, Vol. 7, No. 5 (Nov., 1978), p. 11Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30213415 .

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SCELLANY*MISCELLANY*MISCELLANY*MMISC

Finding a simultaneous solution Previously known as solving simultaneous equations by Tom Leddy, City of Liverpool College of HE

Teachers planning to introduce this topic have a bewildering array of methods to select from. These include use of (i) the intersection of solution sets, (ii) graphs, (iii) matrices, and, at a higher level, (iv) determinants, (v) iterative methods, (vi) Gaussian elimination; and our old friends, (vii) equal coefficients, (viii) substitution.

The choice can be reduced, of course, for such fundamental reasons as "I have never heard of that way" or "My kids can't do matrices". Also a good teacher will want his pupils to meet more than one method. But which is the easiest method to start with?

As Skemp advises, new work is most readily understood if it is easily assimilated into schema which are very familiar to pupils. Thus, when making the choice of method, we should be less concerned with our own preferences and more with the pupils' grasp of earlier work.

This leads us to examine just what previous work is required by -the different methods listed above. In particular let us compare the "equal coefficient" and "substitution" methods.

The "substitution" method involves:

(a) addition and subtraction of the integers,

(b) multiplication of negative integers by natural numbers,

(c) solution of equations of the type Ax= B, where A and B are integers,

(d) algebraic manipulation of the type

2x+ 5y= 2 implies x= (2- 5y)/2,

(e) solution of equations of the type

3(2- 5y) +5y=2, 2

(f) the product of negative integers.

Compare this with the requirements of the "equal coefficient" method. These are simply (a), (b) and (c).

So often, pupils' difficulties with a topic are not inherent in it but arise from dependence on earlier work. The less of this the better.

Calculations in

base thousand by Amos Ehrlich, School of Education, Tel-Aviv University and Science Education Centre, Leeds University

Look at this number: 721 377 051 893

If you think that it is written in base 10, then read it loudly and listen to yourself. Is the way you read it closer to

7x 1011+ 2x 1010+ 1 x 109 + 3x 108+ 7x 107+ 7x 106 + 5 x 104+ 1 x 103 + 8 x 102 +9x10+ 3

or is it closer to

721 x 1 0003+377x 1 0002 + 51 x 1 000+ 893, ie in base 1 000?

Write 051 instead of 51, and our base-thousand numeral will have for "digits" the three-digit numerals of base ten. (In the first "digit" one may omit initial noughts.)

Using an electronic calculator for the multiplication of these base- thousand "digits" (or memorising the 1 000 x 1 000 multiplication table!) one can multiply big numbers as in the following example

2-1- 4434 44

4-2- -

-2" -8 - F a 8--- -

(These "digits" were "carried" in the process.)

721 377 051 893 x 55 487 236

170 244 984 246 748 351 310 624 271 891

39 675 737 854 115

40 027 218 723 371 137 748

Since almost all of the calculators have at least an eight decimal digit display, we can do our multiplication in base ten-thousand:

-2-74

-4-274

7213 7705 1893 x 5548 7236

5219 8843 4749 7748 4002 1998 8390 2364

4002 7218 7233 7113 7748

In the following example of long division, the estimations were done using the - key.

21 197.064 852

17 532 )371 626 941 368 172

3 454 941 3 453 804

1 137.000 1 122 048

14 952 000

Bisecting an angle by Damian Griffin Construction: With centre A and any reasonable radius draw an arc cutting AB at D and AC at F. With centre A and any slightly larger radius draw an arc cutting AB at E and AC at G. Join DG and EF to intersect at H. Join AH. AH now bisects 4 BAC. E B

A H

A F G

Proof: In As AEF, AGD

AE = AG (construction) AF= AD (construction)

,BAC is common

.'. AAEF= AAGD (two sides, included angle)

n As DHE, FHG 4AEF= AdAGD (AAEF -AAGD) 4DHE= FHG (vert. opp. angles)

DE= FG (construction)

.'. ADHE AFHG (two as, corresponding side)

.'. DH=FH 4ADG = 4AFE (AADG-AAFE)

AD= AF (by construction) .. AADH AAFH

(two sides, included angle)

.'. 4DAH= 4FAH

.. AH bisects 4BAC

11

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