Examples. с 1. Reduce - and to equivalent having a common denominatos. с d Here a Xcx d = acd New Hence the fractions hxbx 8 = 'd Numerators. (required are od lcd lcd écd bxcx d = lcd Com, Denom. acd l?d, and 4. Reduce 2r +3 5.7 +1. to a common denominator. 5x2 +2. Ans. and 32 31 6x +9 6. Reduce 7x2 - 1 4.72 to a common denominator. 2.1 2a" Hal Зас x2 1 and to a common denominator. 2r 26 X + 1 7. Reduce 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 23 t har + 2ro to a common denominator. 35 xs Nute. The co-efficient of every term of both numerator and denominator is divisible by 7, and x enters also into every term; + 7x will therefore = 22+ a + 3x ; 2a + 2 + 3 And =5x. Hence, the frac. in its lowest terms, is 57 70 a b 1 8. Reduce Ans, as 69 a” + ab + b 2 Scholium. When, therefore, we have occasion to reduce algebraic fractions to their lowest terms, by finding the greatest 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, siuce 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 fractions required. Examples. 1. To find the sum of g å and f axdxf=adf cxbxf=clf adf cof ebd adf+cof+eba exbxd=ebd Hence and bdf' bdf' bdf bdf bxdxf=ldf =the sum required. and 2.r together. 7 68.ro + 1471 – 35 Ans. 70x 4072 Ans. 154 Ans. 105a + 28a%], + 3062 70ab 169x+77 and Ans. 7 105 37a +116 Ans. 156 Ans. 11x? + 377 + 15 15x + 30 4.x? —78-31 Ans. бr 2.22 2x + 1 Ans. re Ans. 2r2 *-9 Ans. 2a? + 264 a2-bi PROBLEM VI. (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. Examples. 3.7 14.3 1. Subtract from 5 15 C Here 3.2 x 15 = 45% .. 70% 45% 25.7 and 142 X 5 = 70.0 is the Also 5 x 15 = 75 75 75 3 difference required. 2a 4.7 2. From take 36 50 Here (2 a) 5c = 5cx 5ac 5сх. Sac bal 12br and 12a-42)8l=hab 12lir Also 36 50 = 15 lc 1560 15bc 5cr 5ac - hab + 12bx = the difference required. 15ls (45.) To Multiply fractional quantities. Rule. Multiply the numerators together for a new numerator, and also the denominators for a new denominator ; and reduce the products to their lowest term. Eramples. 2r 4.2 1. Multiply by 7 9 Here 28 x 41 = 8.ro , the new Num and 7 X 9 63 . Denom. { : the fraction required is (46.) To divide fractional quantities. Rule. Invert the divisor, and proceed, in all respects, as in Mul. tiplication. (Art. 45.) с = divided by+6 . 15 by by 60 by Erampler. 14.68 2.1 1. Divide 9 3 Invert the divisor and it becomes- ; 22 14.72 3 hence 42 ro X (dividing the numerator and 9 2.r 18r 3 denominator by 6r) is the fraction required. x + a 2. Find the quotient of 2x 2b 51 ta 3 ta 5.r + a Here. 5х2 + баr + a* Х = the quotient 2.2 26 tl 2.0* 26* required. 141 -- 3 10.x 4 70.r 3. Divide Ans. 100 4 20 4. Divide Ans. 63 15ab 150% Ans. 4a 15ab-157,5 6. Divide Ans. 4a' +4ab 4r 12 7. Divide Ans. 5 3 gr 8. Divide Ans. 2 (47.) SCHOLIA 1. If the fractions to be divided have a common denominator take the numerator of the dividend for a new numerator, and the numerator of the divisor for the denominator; for it is evident, that are contained 12 9 as many times in 12 as there are 3's in nine ; i. e. thrice. 2. When a fraction is to be divided by any quantity, it is the same thing whether the numerator be divided by it, or the denoniinator multiplied by it. For, if it be proposed to divide7 by C, we change it into bc, and then dividing the ac by c, we have for the quotient sought. be Hence, when the . is to be divided by c, we have only to x the denominator by 5 5 that quantity, and leave the numerator as it is. Thus • 3 gives -; and 24 9 - 50 16 80° 3. When the two numerators or the two denominators are divisible by a common quantity, we expunge that quantity from each, and use the quotients in place of the fractions first proposed. 4. The signs + and - preserve the same laws in division as in multipli5 3 cation; for X+ and 16 12 1. a ac a fraction is 8 15 3 3 4 |