122. By taking the following instance, we shall see that this rule can be sometimes simplified. Divide ¹⁶/₃₃ by ²⁸/₁₅. Observe that 16 is 4 × 4, and 28 is 4 × 7; 33 is 3 × 11, and 15 is 3 × 5; therefore the two fractions are 4 × 4 4 × 7 ------ and -----, 3 × 11 3 × 5 and their quotient, according to the rule, is 4 × 4 × 3 × 5 --------------, 3 × 11 × 4 × 7 in which 4 × 3 is found both in the numerator and denominator. The fraction is therefore (108) the same as 4 × 5 20 ------, or ----. 11 × 7 77 The rule of the last article, therefore, admits of this modification: If the two numerators or the two denominators have a common measure, divide by that common measure, and use the quotients instead of the dividends. 123. In dividing a fraction by a whole number, for example, ⅔ by 15, consider 15 as the fraction ¹⁵/₁. The rule gives ²/⁴⁵ as the quotient. Therefore, to divide a fraction by a whole number, multiply the denominator by that whole number. EXERCISES. Dividend. Divisor. Quotient. 41 63 41 ---- ---- ----- 33 11 189 467 907 47167 ---- ---- ------- 151 101 136957 7813 601 13 ----- ---- ---- 5071 11 461 ¹/₅ × ¹/₅ × ¹/₅ - ²/₁₇× ²/₁₇ × ²/₁₇ What are -----------------------------------, ¹/₅ - ²/₁₇ and ⁸/₁₁ × ⁸/₁₁ - ³/₁₁ × ³/₁₁ ----------------------- ? ⁸/₁₁ - ³/₁₁ 559 _Answer_, ----, and 1. 7225 A can reap a field in 12 days, B in 6, and C in 4 days; in what time can they all do it together?[16]--_Answer_, 2 days. [16] The method of solving this and the following question may be shewn thus: If the number of days in which each could reap the field is given, the part which each could do in a day by himself can be found, and thence the part which all could do together; this being known, the number of days which it would take all to do the whole can be found. In what time would a cistern be filled by cocks which would separately fill it in 12, 11, 10, and 9 hours?--_Answer_, (2⁴⁵⁴/₇₆₃) hours. 124. The principal results of this section may be exhibited algebraically as follows; let _a_, _b_, _c_, &c. stand for any whole numbers. Then _a_ 1 _a_ _ma_ (107) ---- = ---- × _a_ (108) ---- = ---- _b_ _b_ _b_ _mb_ _a_ _c_ _ad_ _bc_ (111) --- and --- are the same as ---- and ---- _b_ _d_ _bd_ _bd_ _a_ _b_ _a_ + _b_ (112) --- + --- = --------- _c_ _c_ _c_ _a_ _b_ _a_ - _b_ --- - --- = --------- _c_ _c_ _c_