Ex. 7. Required the greatest common divisor of (da -c) Xaa+c-dc2 and 4da” -(2c +4cd)at 2c2. Arranging these quantities according to d, we have (a? .ca)da +c4-a2c, or (ao ---ca)d’ –(a2-co)c, and (4a2 —4ac) xd-(ac) x 2c? '; it is evident, by inspection only, that a? —co is a divisor of the first, and a-c of the second. But a--cis divisible by a-C; therefore a-c is a divisor of the two proposed quantities : Dividing both the one and the other by amc, the quotients will be (a+c) (dc), and 4ad-2c?; 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* --<4 and y3 --yox--yo+x". Ans. yo — 32. Ex. 9. Required the greatest common divisor of a-64 and a6-68. Ans. a -62. Ex. 10. Required the greatest common divisor of at ta'b-ab3-64 and a' t-aob+64. Ans. a' tab +6. Ex. 11. Required the greatest common divisor of 4--2ax + x? and a 3-ax-ax2 + x3. Ans. a--X. Ex. 12. Find the greatest common divisor of 623 ---Byx2 + 2yo x and 12x2--15yx+3y. Ans. x-y. 2 Ex. 13. Find the greatest common divisor of 36b'a -1862a5 - 27ba" + 9ba3 and 27bas1862a* -9ba. Ans. 96?a* --962a3. Ex. 14. Find the greatest common divisor of (c-d)a2 +(2bc-2bd)a+(6ac-62d) and (bc-bd+ c-cd)a+(6d+bc-bc-bcd). Ans. Cd. c-. Ex. 15. Find the greatest common divisor of qnp3 +3npoq2 - 2npq: ---2nq* and 2mpaq – 4mp4 -mp3q+3mp3. 3 2 Ans. q-p Ex. 16. Find the greatest common divisor of 23 +9x2 +-272--98 and x2 +1 2.r -- 23. Ans. * 2. § III. METHOD OF 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 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 quantitics, whose greatest common measure, or divisor, is x, and let a=mx, b=nx; 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; mno is, therefore, the least common multiple of a and b. Now mx=a, and nx=b; therefore mw Xnx=aXb, 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; 12 X 18 therefore their least common multipleis- = 36. Every other common multiple of a and b is a multiple of mna. Let q be any other common multiple of the two quantities, and, if possible, let mnt be con 6 tained in q, r times, with a remainder s, which is less than mnx; then q-rmnr=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 mnr, 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 u be the greatest common divisor of a and b; ab y of a and c. abc of X and d, xy &c. &c. ab then will be the least common multiplier of a and b: abc of a, b, and c: wy abcd of a, b, c, and d: my? &c. &c. 147. Hence the following method for finding the least common multiple of two or more quantities. RULE. I. 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. III. 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 a3b2x, acbx, and abco d. Here, the greatest common measure of a3b2 x and acbx, is abx, and the least common multiple is, a3b2 x Xacbx therefore, =a3b2cæ; the greatest abx common measure of a’bocx2 and abcad is abc; a'baco Xabcd hence, =a3b2 c dx2 = the least mul abc tiple required. Ex. 2. Required the least common multiple of 2a2x, 4ax, and 6x3. Here, the greatest common measure of 2a x and 4axo, is 2ax; hence, the least common multiple is 2ax X 40x2 -4a2x2 ; again, the greatest common 2ax measure of 4ao x2 and 6x3 is 2x2 ; and therefore 4a2x2 X 63 1:la? x3 = the least common multiple 2xca required. * Ex. 3. Required the least common multiple of 3a, 4a", and 12ab. Ans. 24a2b. Ex. 4. Required the least common multiple of ao-b, a+b, and a’ +6. Ans. a* -54. Ex. 5. Required the least common multiple of 27a, 156, 9ab, and 3a2. Ans. 135a2b. Ex. 6. Required the least common multiple of a' + 3aPb+3ab2 +6°, a' + 2ab +62, a -67. Ans. a*-+-2a3b-2ab3--64. Ex. 7. Required the least common multiple of a+b, a--b, a2 + ab +6", and a? -ab-+-63. Ans. a 6-62. § IV. REDUCTION OF ALGEBRAIC FRACTIONS. CASE I. To reduce a mixed quantity to an improper fruction. 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. 27 Ex. 1. Reduce 3x+5 . - lo an improper fraction. ба The integral part 3x multiplied by the denominalor 5a of the fraction plus the numerator (26) is equal to 3x X 5a +21=15ax +2b: 15ax-4-26 Hence, is the fraction required. 50 |