result by each of these several factors except ğ, employing cancellation in the division wherever practicable. 9. Multiply 6 by 43; 343 by 6; 27 by 8; 884 by. It is usual to change mixed numbers into improper fractions before multiplying them. But, in some of these, as well as in many other cases, it will be found quite as easy, and much shorter, to multiply without any such change. For example, But in division, where the divisor is a mixed number, and the dividend is either a mixed number or an integer, it will be found most convenient to change both to the form of improper fractions. 10. Divide 32 by 43; also 16 by 52; and prove by multiplication. RAPID AND CONCISE METHODS OF COMPUTING WITH COMMON FRACTIONS. In all treatises on arithmetic, the pupil is directed to bring fractions that are to be added or subtracted to a common denominator. But, when the fractions do not exceed two in number, these operations can frequently be performed much more rapidly and quite as correctly by bringing them to a common numerator, as will appear from the following exemplifications and exercises: I.—Addition by a Common Numerator. Exemplification for the Black-board. 1. Find the sum of and . (7+3)=(35+3)= }} . Suggestive Questions.-1. Compare the new numerator with the given denominators, and say what relation it bears to them. Is it their sum, difference, product, or quotient? The given numerators remaining the same (1), would the new numerator be the sum of the given denominators, whatever might be their numbers? 2. What relation does the new denominator bear to the given denominators; is it their sum, difference, product, or quotient? Would it be so whatever were the numbers of the given denominators? How, then, can any two fractions of different denominations be added by simple inspection if their numerators be 1 ? 2. Add and by inspection; that is, by using the sum and product of their denominators for the sum of the two fractions, and prove by addition in the old method. 3. Add, in the same manner, and 4; +; }+}; ++ +; and prove each by addition in the old method. Suggestive Questions.-If the sum of 1 seventh and 1 fifth be 12 thirty-fifths, what will be the sum of 4 sevenths and 4 fifths; that is, how many times will the amount be greater than the other? How many times will the sum of 3 sevenths and 3 fifths be greater than the sum of 1 seventh and 1 fifth? How, then, can you add two fractions by inspection, when their equal numerators are greater than 1? 4. Add, by inspection, and [1×4=§§], or, omitting superfluous figures [+]. 5. Add and by inspection, and prove by addition in the old method. 6. Add and prove in the same manner, and; ‡ and and ; and ; and ; and 5. Remark.-When both terms are unlike in the fractions to be added, a single glance will generally show whether it be easier to make the numerators or denominators common; and, by having a choice, it will very rarely be necessary to change both fractions before they are added. The numerators may be altered as follows: (+9)=(21+), by multiplication; (+ 3)=(+14); (5+15)=(1+1) by division. In all these cases, it will be perceived that it is a more simple operation to change the numerators than the denominators. 7. Add, by inspection, and §, and prove by the old method. 8. Add, by inspection, and ; and ing numerators to 6); and ; and (chang ; and % and 18. II.-Subtraction by a Common Numerator. [Subtraction and Addition by this method differ in no respect, save that the difference, instead of the sum of the given denominators, constitutes the new numerator.] 1. Find the difference of and by inspection, and prove by the old method. 2. Find the difference of and, as above, and prove. 3. Find the difference of and 4, as above, and prove. 4. Find the difference of as above, and prove. and ; and ; and ; III.-Multiplication. Where the numerator of one factor is equal to the denominator of the other. Exemplification for the Black-board. 1. What are the several products of by, and by &, in their lowest denominations? Suggestive Questions.- Which of the numbers in the two given factors are retained in the product; the equal or the unequal? Do the unequal numbers retain their respective places, or are they reversed in the product? 2. Multiply by by striking out the equal numbers, and prove by division. 3. Multiply by, by, by far by 1, by inspection, and prove by division. When all the numbers in the two factors are unequal. Exemplification for the Black-board. 1. Multiply by, by 1, 4 prove by division. Suggestive Questions.-How can with 6 for a numerator? How can with 5 for a numerator? How can with 3 for a denominator? 17 by 7, by inspection, and be changed to a fraction be changed to a fraction be changed to a fraction 2. Multiply by 19, by, by, by, by inspec tion, and prove by division. IV.-Division. Where the two numerators or the two denominators are 1. Divide, in the usual manner, by 3, and 4 by . Suggestive Questions -Comparing the numbers in the given divisor and dividend of both problems with those of the quotients: which are removed, those that are equal, or those that are unequal? Where is the remaining number of the divisor found in the quotient? Ans. In the vacant place in the 2. Divide by by inspection, and prove by multiplication by inspection. 3. Divide by, by 8, it by 14, by 7, 8 by 31, 15 by, all by inspection, and prove as above. Where all the given terms are unequal. 1. Divide by, by 12, 8 by §, by inspection, and prove by multiplication. Suggestive Questions.-How can & be changed to an equivalent fraction, with 6 for a numerator? How can be changed to an equivalent fraction, with 7 for a denominator? How can be changed to an equivalent fraction, with 5 for a numerator? 8 2. Divide by 17, by 24, prove as above. 3. Divide by, by, prove. by, by inspection, and by, by inspection, and V.-Multiplication and Division by Addition or Subtraction. Exemplifications for the Black-board. 1. Multiply 32 by 15. Suggestive Questions.-If 32 were to be multiplied by 18 (=1) instead of 18, how much too large would the number be? Ans. too large. Then what portion of 32 must be subtracted to make the process correct? 32×18=32X18 (or 1)-; therefore, 32x+8=32 32X2 or, omitting superfluous figures, 32X1=30. 32X+8=30 2. Multiply 54 by 12. of 54-74; therefore, 5. Multiply each of the following numbers, namely, 216, 325, 84, 125, 64, 236, 27, §, 7, 9, severally by 11, by 24, by 23, by, by 8, by, by g, by 14, by 13, by, by 7, by g, by 1, by 26. As the last three factors are improper fractions, the process consists of addition. Why? Try the problem, and see. Prove each of the above by multiplying in the usual manner. Exemplifications for the Black-board. 6. Divide 42 by 7. 42x=42+4 of 42-48. 37x=37+ of 37-47. 242=24+1 of 24-223. 24X9-24- of 24-204. |