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by the same reduction, become respectively

adfh cbfh ebdh gbdf

bdfh' bdfh' b d fh' b d f h

52. I have given, in art. 79 of arithmetic, a process for obtaining, in certain cases, a denominator more simple, than that which results from the general rule; it may be much simplified by means of algebraic symbols, as we shall see.

If, for example, we have the two fractions

a d

it is

to

easy bc bf'

see that the two denominators would be the same, if ƒ were a factor of the first, and c a factor of the second; we multiply then the two terms of the first fraction by f, and the two terms of the

second by c, which gives

af cd
and
bcf bcf'

abf

more simple than

bbc f

bcd

and

bbc f'

obtained by multiplying by the original denominators.

In general, to form the common denominator, we collect into one product all the different factors raised to the highest power found in the denominators of the proposed fractions; and it remains only to multiply the numerator of each fraction by the factors of this product, which are wanting in the denominator of the fraction.

Having, for example, the fractions

a d

e

and I b2 c' bf' cg

form the

product b2 cfg; I multiply the numerator of the first fraction by fg, that of the second by b c g, that of the third by ba f, and I

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The whole of a added to the fraction, or the expression

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d

d

d

b

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C

ac + b

с

+

ac

b

b

=

+

=

+

C

с

ac b ac b

с

=

C

C

с

The terms of the preceding fractions were simple quantities, but if we had fractions, the terms of which were polynomials, we should have to perform, by the rules given for compound quantities, the operations indicated upon simple quantities; it is thus that we have

a2 + b2 a- -b (a2 + b2) (a—b)

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Ɑ- -b (c+d) (a-b) ac+ad-bc-bd'

and so of other operations.

54. Understanding what precedes, we can resolve an equation. of the first degree, however complicated.

If we have, for example, the equation

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we begin by making the denominators to disappear, indicating only the operations; it becomes then

(a+b)(x−c)(3a+b)+4b(a--b)(3a+b)=2x(a−b)(3a+b)-—ac(a--b); performing the multiplications, we have 3a2x+4abx+b2x-3a2c-4abc-b2c+12a2b-Sab3-4b3 = 6a2 x- - 4 a b x 2 b2 x — - a2c + abc;

transposing to one member all the terms involving x, it becomes -3a3x+8abx+3b2x=2a2c+5abc+b2c-12ab+8ab2+4b3,

from which we deduce

x=

2a2c+5abc + b2 c-12 a2 b + 8 a b2+4b3

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Of Questions having two Unknown Quantities, and of Negative Quantities.

55. THE questions, which we have hitherto considered, involve only one unknown quantity, by means of which, with the known quantities, are expressed all the conditions of the question. It is often more convenient, in some questions, to employ two unknown quantities, but then there must be, either expressed or implied, two conditions, in order to form two equations, without which the two unknown quantities cannot be determined at the same time. The question in art. 3, especially as it is enunciated in art. 4, presents itself naturally with two unknown quantities, that is, with both the numbers sought. Indeed, if we denote

the least by x,
the greatest by y,

their sum by a,

their difference by b,

we have, by the enunciation of the question,

x + y = a,

y x = b.

Each of these two equations being considered by itself, we can determine one of the unknown quantities. If we take the second, for example, we deduce the value of y, which is

y = b + x,

a value, which seems at first to teach us nothing with regard to what we are seeking, since it contains the quantity x, which is not given; but if, instead of the unknown quantity y in the first equation, we put this, its equivalent; the equation, containing now only one unknown quantity x, will give the value of x by the process already taught.

We have in fact by this substitution,

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and putting this value of x in the expression for y,

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we have then for the two unknown numbers the same expressions as in art. 3.

It is easy to see indeed, that the above solution does not differ essentially from that of art. 3; only I have supposed and resolved the second equation y xb, which I contented myself with enunciating in common language in the article cited; and from it I deduced, without algebraic calculation, that the greater number was a + b.

56. I take another question.

A labourer having worked for a person 12 days, and having with him, during the first 7 days, his wife and son, received 74 francs; he worked afterward with the same person 8 days more, during 5 of which, he had with him his wife and son, and he received at this time 50 francs; how much did he earn per day himself, and how much did his wife and son earn?

Let x be the daily wages of the man,

y that of his wife and son;

12 days' work of the man will amount to 12x,

7 days' work of his wife and son,

7y;

we have then by the first statement of the question,

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Proceeding as in the preceding question, we take the value of y in the first equation, which is

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and substitute this value in the second, multiplying it by 5, the coefficient, and it becomes

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an equation, which contains only the unknown quantity x. By reducing it, we have

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transposing -4 x to the second member, and 350 to the first,

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Knowing x, which we have just found equal to 5, if we place this value in the formula

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the second member will be determined, for we have

74-12 X 5 74-60 14

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= = 2;

7

7

7

y = 2.

The man then earned 5 francs per day, while his wife and son earned only 2.

57. The reader has perhaps observed, that in resolving the above equation 370 4 x 350, I have transposed 4x to the second member. I have proceeded thus to avoid a slight difficulty, that would otherwise have occurred, and which I will now explain.

By leaving 4 x in the first member, and transposing 370 to the second, we have

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and reducing the second according to the rule in art. 19, there will result from it

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But as we have avoided, in the preceding article, the sign, which affects the quantity 4 x, by transposing this quantity to the other member; and as in like manner the quantity 350-370 becomes by transposition 370- 350; and since a quantity, by being thus transferred from one member to the other, changes the sign (10), it is evident that we may come to the same result by simply changing the sign of each of the quantities — 4 x +350 370, which gives

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