Eramples. 2. Divide a +5a'r +100°r* +- 10a®z + bart tr. Ans. a* + 3 a 1 + 3 ars + . 3. Divide 12 x 13 r* — 34 x' + 35 x* by 4.m.; - 78. Ans. 3r + 2r2 5r. 4. Divide z' — t - H2 + 2r - 1 by x? + j. Ans. It m3 + x2 + I. 5. Divide yo — 3y*x® + 3yor* + zo by ys – 3yor + 3yr* — 23. Ans. y' + 3yor + 3yrs + II 6. Divide as 5a'x + Joao.ro 10a’r + 5art -- XS by a' Zar + 3*. Ans. a- За'r + Зах? . Miscellancous Examples. 1. Divide as I by a — x. Ans, a' ta'r -- a* ra t ar + r. 2. Divide 4.23 76ax: + 35a’r + 105a2 by 2r 3a. Ans. 2x 35ar - 35a?. 3. Divide 23 + y by r + y. Ans. x2 y? - yr. 4. Divide 2a* 32 by a Ans. 2a3 + 4a2 + 8a + 16. 5. Divide as + air •a2x2 7a-r! + 6x by a? 7. Ans, a' t a? - 6r. Divide x + 3r2y + 3ry? t ys by za + 2xy + ye. Ans. x + y. 7. Divide a' — 3a?c + 4ac - 26 by a? — 2ac + c. acc Ans. a 6+ a 2ac + c 8. Divide 6.x* — 96 by 3r – 6. Ans. 2x3 + 4x2 + 80 + 16. 9. Divide 1 50 + 10x2 - 10.23 + 5rt Is by 1 -2x + xs. Ans. I 30 + 3.2% - 13. 2. FRACTIONS. Algebraic Fractions are subject to the same rules as are the fractions of commun arithmetic, and their reduction requires only the application of the principles laid down in the preceding rules. PROBLEM I. Rule. Multiply the integer by the denominator of the fraction, and to the product annex the numerator with its proper sign ; under this same place the former denominator, and the result will give the improper fraction required. Example 1. 2.0 Reduce 3a + to an improper fraction. Here the integral part x denom, of the fraction + the numerato 3a X 5a+ 202 = 15a + 2x. Hence is the fraction required. 5a 15 a + 2x The one to an improper fraction C 2. Reduce 51 30 apk 4.2 Ans. ба? 23 3. Reduce 57 to a common denominator. 330 + 3* Ans. 7 20 4. Reduce hab + to a common denominator. За 40 5. Reduce 362 to a common denominator. 5.3 a 6. Reduce a It to a common denominator. C a PROBLEM II. Rule. Observe which terms of the numerator are divisible by the denominator without a remainder, the quotient will give the integral part ; to this annex the remaining terms of the numerator, with the denominator under them, with their proper signs, and the result will be the mixed quantity required. Erample 1. Reduce a2 + ab + be to a whole or mixed quantity. Here az + ab 62 a, 23 2. Reduce 30 Ans. 3a + 5a 5a 5a sa 3. Reduce 2013 12a + 40 3c 4. Reduce 42 10x2 2/8 5. Reduce a 42% Ans. 2x + 3.20 PROBLEM III. Rule. Multiply each numerator into every denominator 'ut its own for the new numerators, and multiply all the denominators for a common denominator. Sa mukiplied by 7 = 353 ; and 28 - 3 subtracted from 35x leaves 53378 i a Examples and , to equivalent having a common denominator. . с 1. Reduce с 7 d Here a Xcx d = acd New Hence the fractions bi xbxd=l'd Numerators. required are схих acdb2d, c'd and xcx d = bod Com. Denom. lcd lcd écd C 3 cd 6x +9 5x +2. 3r 31 5. Reduce 41+2r 372 2r and to a com. denominator. 5 4a 3ს 45 biza 40 ar and 60 ab 60 al 60 ab 6. Reduce 1 to a common denominator. 2x 2atal Зас 7. Reduce X2 and 27 26 to a common denominator. X + 7x2 PROBLEM IV. To reduce a praction to its lowest terms. Rule. Observe what quantity will divide all the terms of both mumerator and denominator without any remainder : divide them by this quantity, and the fraction is reduced to its lowest terms. Examples. 1. Reduce 14 2 + 7ar + 2x to a common denominator. 35 Nute.-The co-efficient of every term of both numerator and denominator is divisible by 7, and x enters also joto every terın; + 7x will therefore divide both nomerator and denominator without a remainder. For 14./9 + 7ar + 21x2 = 22a + a + 3x ; 35 ro 2a + a +3 And =5x. Hence, the frac. in its lowest terms, is 77 5x 72 2. Reduce 0 to its lowest terms. 12 a? 1 Ans. a to a b 1 8. Reduce Ans, a + ab + 1,2 Scholium. When, therefore, we have occasion to reduce algebraic fractions to their lowest terms, by finding the greutest cuinmon meusure of the numerator and the denominator, as in conimon arithmetic, it is evident that the rule which we have given for the solution of the foregoing problem answers every practi. cal purpose, since it is only requisite to discover what quantity will divide both numerator and denominator, and that the quantity must therefore be the greatest common measure. PROBLEM V. (43.) To add fractional quantities. Rule. Reduce the fractions to a common denominator, as in Problem III. Add all the numerators together, and below their sum set down the common denominator, and it will give the sum of the frac tions required. +3, 3r_1 and 2r + 3 4.0 2. Add and together. 7 703 4072 Ans. 154 31 Ans. 105a + 28a"), + 306 4. Add 26 70ab 169x+77 5. Add and Ans. 105 and 6. Add 37a +116 Ans. 150 11x + 377 + 15 and Ans. 15x + 30 4.3% -72-31 Ans. Or 2x + 1 XP 7. Add 9. Add (44.) To subtract fractional quantities. Rule. Reduce the fractions to a common denominator as in Problem III ; and then subtract the numerators from each other, and under the difference, write the common denominator. Eramples. 3.7 14.0 1. Subtract (rom 5 15 C Here 32 x 15 = 45% .:: 70x 457 25.7 and 147 x 5 = 70.7 is the Also 5 * 15 = 75 75 75 3 difference required. 2a 4.2 2. From take 30 50 Here (Z - a) 5c = 50% 5ас 5cx sac bal 12br and (2am-fx)8l=6ab 12lir Also 36 50 = 15 lc 15bc 15bc 501 5ac 6ab + 12bx the difference required. 15ls |