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Ex. 7. Required the greatest common divisor of (da -c) Xa+c-doc? and 4da-(2c +4cd)at 2c?. Arranging these quantities according to d, we have (a? ca)da +c4-a2c>, or (ao ---c?)d(az-co)c”,

and (4a’ — Hac) xd-(a--c) x 2co; it is evident, by inspection only, that a’ —c? is a divisor of the first, and a-c of the second. But a? ---cis divisible by arc; therefore a-c is a divisor of the two proposed quantities : Dividing both the one and the other by a—c, the quotients will be

(a+c) (d-c), and 4ad2c2; which, by inspection, are found to have no common divisor, consequently a-c is the greatest common divisor of the proposed quantities.

Ex. 8. Required the greatest common divisor of y* --x4 and yo-y-x--yx +2%. Ans. y' 32.

Ex. 9. Required the greatest common divisor of a 64 and a -68.

Ans. ai_62. Ex. 10. Required the greatest common divisor of at ta'b-a63-64 and a ta’62 +64.

Ans. a” tab +6. Ex. 11. Required the greatest common divisor of 02_-2ax + x2 and a 3-ax-ax2 + x3.

Ans. a-X. Ex. 12. Find the greatest common divisor of 6x3---Byx2 + 2yo x and 12x*--15yx+3y2.

Ans. 2-y.

Ex. 13. Find the greatest common divisor of 36b2a6-18625 - 2762a4 + 9b2a3 and 2763a5. 1862a4-9bas.

Ans. 9b2a4-9b'a3. Ex. 14. Find the greatest common divisor of (c-da? +(2bc-2bd)a +(62c-62d) and (be-bd+ c-cd)a+(62d+bc-b2c-bcd). Ans. c-d.

Ex. 15. Find the greatest common divisor of anp3 + 3npoqa —2npq: --2nge and amp?q*__mp4 -mpq+3mpq3

Ans. 9.

Ex. 16. Find the greatest common divisor of 73 +9x2 +278--98 and ro+12.7--23.

Ans. x -2.

§ III. METHOD Or FINDING THE LEAST COMMON MUL

TIPLE OF TWO OR MORE QUANTITIES.

145. The least common multiple of two or more quantities is the least quantity in which each of them is contained without a remainder. Thus, 20abc is the least common multiple of 5a, 4ac, and 2.

146. The least common multiple of two numbers, or quantities, is equal to their product divided by their greatest common measure, or divisor,

For, lel a and b be any two quantities, whose greatest common measure, or divisor, is t, and let a=mx,b=ng; then mnx is a multiple of a by the units in n, and of b by the units in m; consequently it is a common multiple of a and b.

But since x is the greatest common measure of a and b, m and n can have no common divisor; mnx is, therefore, the least common multiple of a and b. Now mx=d, and nx=b; therefore mw Xnx=a Xb,

ab (Art. 50), and mnx=-,(Art. 51). Hence the rule is evident; as for example:

Let the least common multiple of 18 and 12 be required. Their greatest common measure is 6

i

12 X 18 therefore their least common multipleis =36.

6 Every other common multiple of a and b is a mul. tiple of mna.

Let q be any other common multiple of the two quantities, and, if possible, let mnr be con

tained in q, r times, with a remainder s, which is less than mnx ; then q--rmnx=s; and since a and b measure q and rmnx, they measure q-rmnx, or 3, (Art. 131); that is, they have a common multiple less than mnx, which is contrary to the supposition.

To find the least common multiple of three quantities a, b, and c. Let m be the least common multiple of a and b, and n the least common multiple of mand c; then n is the least common multiple sought. For every common multiple of a and b is a multiple of m, therefore every common multiple of a, b, and c, is a multiple of m and c; also, every multiple of m and c is a multiple of a, b, and c; consequently the least common multiple of m and c is the least common multiple of a, b, and c. Or, in general, let a, b, c, d, &c. be any set of quantities, and let be the greatest common divisor of a and b;

ab y

and c, abc

and d.

ху &c.

&c. ab then will be the least common multiplier of a and b: abc

of a, b, and c: ху abcd

of a, b, c, and d: ry &c.

&c.

of

of

147. Hence the following method for finding the least common multiple of two or more quantities.

RULE.

1. Find the least common multiple of two quantities, by the preceding Article.

II. Find, in like manner, the least common multiple of the result, thus found, and the third quantity.

Ill. Find also the least common multiple of the last result and the fourth quantity; and proceed in the same manner with this result and the fifth; and so on: the result last found will be the least common multiple of all the quantities.

Ex. 1. Required the least common multiple of a3b2, acbx, and abcd.

Here, the greatest common measure of a3b2 x and acbx, is abx, and the least common multiple is,

a3b2 x Xacbx? therefore,

=a3b2cx ? abx

the greatest

; common measure of a3b2c«2 and abcd is abc;

a3b2cc2 x abcd hence,

=a3b2c2 d.x2 =the least mul

abc tiple required.

Ex. 2. Required the least common multiple of 2a2x, 4axo, and 6x 3.

Here, the greatest common measure of 2a x and 4ax", is 2ax; hence, the least common multiple is 2a2x X 4axa

=4a2x2 ; again, the greatest common 2ax measure of 4a2x2 and 6x3 is 2002 ; and therefore 4a2x2 X 6x3

-=12a.co= the least common multiple 2x2 required.

*Ex. 3. Required the least common multiple of 3a, 4a", and i 2ab.

Ans. 24aob. Ex. 4. Required the least common multiple of ao-b, atb, and a2 +6".

Ans. af-b4. Ex. 5. Required the least common multiple of 27a, 156, 9ab, and 3a2.

Ans. 135aab. Ex. 6. Required the least common multiple of a' + 3aPb+3ab? +6', a' +-2ab +b?, a--62.

Ans. a4 +-2a3b-2ab3 --64. Ex. 7. Required the least common multiple of atb, a---b, a2 + ab +6', and aa-ab-t-63.

Ans. a6-62.

$ IV. REDUCTION OF ALGEBRAIC FRACTIONS.

CASE I.

To reduce a mixed quantily to an improper fraction.

RULE.

148 Multiply the integral part by the denominator of the fraction, and to the product annex the numerator with its proper sign: under this sum place the former denominator, and the result is the improper fraction required.

211 Ex. 1. Reduce 3x+to an improper fraction.

5a The integral part 3x multiplied by the denominator 5a of the fraction plus the numerator (26) is equal to 3x X 50 +2=15ax+ 2h :

15ax 4-26 Hence,

is the fraction required.

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