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EXAMPLES.

1. Required the L. C. M. of the quantities 6a2+11ax +3x2 and 6a2+7ax-3x2.

The G. C. M. of these quantities is 2a+3x, (Exercise 3, Art. 60,) hence the quotient of the first quantity and this measure is 3a+x; then (6a2+7ax-3x2) (3a+x)=18a3 +27a2x—2ax2-3x3 is the L. C. M. required.

2. Required the L. C. M. of 325 and 455.

The G. C. M. of these numbers is 65; then 325÷÷÷65 =5; therefore 455×5=2275 is the L. C. M. required.

EXERCISES.

1. Find the L. C. M. of 8x2, 12x3, and 20x1·

Ans. 120x4.

2. Find the L. C. M. of x-1, x+1, x2—1.

Ans. x2-1.

3. Find the L. C. M. of 6ax, 15bx, 27b, and 35abx. Ans. 1890abx.

4. Find the L. C. M. of the nine digits. Ans. 2520. 5. Find the L. C. M. of x3-a3 and x2—a2.

Ans. x4+ax3—a3x—aa.

6. Find the L. C. M. of x3-3x2+7x-21 and xa—49, Ans. x-3x-49x+147.

7. Find the L. C. M. of a4 —x1 and a3—a2x—ax2+x3. Ans. a5— aax—ɑxa+x5. 65. PROP. I. A fraction is multiplied by any quantity when its numerator is multipled, or its denominator divided, by that quantity.

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a

be the fraction, and x its value,* so that-x;

* This is merely expressing the result of the division of a by b,

by the simple symbol a, instead of the complex symbol.

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a+a

= ; but, by the nature of division,

b

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a+a

b

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to n terms =x+x+x &c. to n terms =nx, therefore n

α

b

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a

bn

a

Again, let the fraction be and its value, so that

bn'

=x; then, by the nature of division, a=bnx, and

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(Art. 30) == nx; but is the form which the original

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fraction assumes, when its denominator is divided by n: therefore, the division of the denominator multiplies the original value of the fraction.

66. II. A fraction is divided by any quantity when its numerator is divided, or its denominator multiplied, by that quantity.

an

Let the fraction be: this (Art. 65) is the product

a

which arises from multiplying the fraction by n; if,

b

therefore, this product be divided by one of the factors n,

H

the quotient will be the other factor - : hence,

a

an

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a

b

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be the fraction, and x its value, so that

= a : if the denominator be multiplied by n, the frac

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tion becomes ; but to divide by a product is the same

nb

as to divide by its factors in succession, therefore

a

nb

= (a÷b)÷n: now, a÷b = = =x, therefore (a÷b)

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67. III. If the terms of a fraction be multiplied or divided by the same quantity, the value of the fraction is not changed.

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Let be the fraction, and x its value, so that

b

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then, if the numerator be multiplied by n, the fraction

will become

an

=nx (Art. 65); and if the denominator of

this new fraction be multiplied by n, we shall have

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an

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bn

be the fraction, and x its value, so that

= a: then, if the denominator be divided by n, the

an

bn fraction will become -=na (Art. 65); and if the nume

α

get=

rator of this new fraction be divided by n, we get = x

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68. PROB. I. To reduce a fraction to its lowest terms. Find the greatest common measure of the terms (Art. 58), and divide both the terms by it.*

It follows from the nature of the greatest common measure, that when both terms of a fraction are divided by it, they are rendered prime to one another, and the fraction is therefore irreducible; the reduced fraction is of the same value as the original one by Art. 67.

EXAMPLES.

24a2x3y2

1. Reduce

to its lowest terms.

8abx1

Here 8ax3 is plainly the highest divisor of both terms;

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Here a+ will obviously divide both terms: dividing,

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* Or, find by inspection any common divisor of the terms, and divide them both by it; the fraction is thus reduced to lower terms: repeat the process till no common divisor of the terms can be found, and the fraction will then be in its lowest terms.

Here, since 16 and 6 are the doubles, and 24 and 9 the

triples of 8 and 3, it is plain that

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The greatest common measure must, in this case, be found

by Art. 58. The work is as follows:

2763-12-28

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Whence the last divisor, 3x+7, is the greatest common measure; dividing both terms of the original fraction by it, we obtain, for the reduced fraction,

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