by the same reduction, become respectively adfh cbfh ebdh gbdf bdfh' bdfh' b d f h' b d f h 52. I have given, in art. 79. of arithmetic, a process for obtaining, in certain cases, a denominator more simple, than that which results from the general rule; it may be much simplified by means of algebraic symbols, as we shall see. If, for example, we have the two fractions a d bc bf' it is easy to see that the two denominators would be the same, if ƒ were a factor of the first, and c a factor of the second; we multiply then the two terms of the first fraction by f, and the two terms of the second by c, which gives and and af bcf cd more simple than bcf' abf b b c f obtained by multiplying by the original denominators. b c d b b c f In general, to form the common denominator, we collect into one product all the different factors raised to the highest power found in the denominators of the proposed fractions; and it remains only to `multiply the numerator of each fraction by the factors of this product, which are wanting in the denominator of the fraction. Having, for example, the fractions a d e and I form the b2 c' bf" cg' product befg; I multiply the numerator of the first fraction by fg, that of the second by bcg, that of the third by b2ƒ, and I obtain afg bcdg b2 ef b2 cfg' b2 cfg' b2 cfg The terms of the preceding fractions were simple quantities; but if we had fractions, the terms of which were polynomials, we should have to perform, by the rules given for compound quantities, the operations indicated upon simple quantities; it is thus that we have 54. Understanding what precedes, we can resolve an equation of the first degree, however complicated. If we have, for example, the equation we begin by making the denominators to disappear, indicating only the operations; it becomes then (a+b)(x−c)(3a+b)+4b(a−b)(3a+b)=2x(a−b)(3a+b)—ac(a−b); performing the multiplications, we have 3a2x+4abx+b2x-3a2c-4abc-b2c+12a2b8ab2-463 = 6a2 x — 4 a b x 2 b2 x — a2c + abc; transposing to one member all the terms involving x, it becomes -3a2x+8abx+3b2x=2a2c+5abc+b2c—12ab+8ab2+463, from which we deduce x= 2a2c+5abcb2c-12 a2b+8ab2+463 -3a28ab +362 Examples in Division in which the Divisor is not an Aliquot Part of the Dividend. 1. 1 ÷ (1 − 6) = 1 + b + b2 + b3 +64 + 2. 1(1+6) = 1 − b + b2 — b3 + b4 3. c ÷ (a — b) = -1/2 + Examples in the Reduction of Fractions. 3 a 1.88 ++h= 56 4 d e g 12 ad +5 b c + 20 b d h F k 20 b d adfhbcfh-bdeh-bdfg-bdfhk 5. c2ab3ac (ab) efgdg-cf b2c5ab2 c + a3 |