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150. Other Changes in the Terms of a Fraction. 1. Reduced to an equivalent fraction, having 6 for a numerator.

Solution. — Observing that the proposed numerator, 6, is two times the given numerator, 3, we multiply both terms by 2, which gives

3 X 2

4 X 2 6

3 8' or we may at once write

4 2. Reduce f to an equivalent fraction, having 10 for its numerator.

10 is

of

6

8

or

5

8 or

9

6

the given numerator, 8, we multiply both terms by or by lị, which

4 8 8 X 11 10

10 gives 9 9 X 17

113 111 3. Reduce to an equivalent fraction, having 10 for its numerator.

4. Reduce to an equivalent fraction, having 9 for its numerator.

5. Reduce to an equivalent fraction, having 6 for its numerator.

6. Reduce 4 to twenty-firsts.

Solution. — Observing that the proposed denominator, 21, is three times the given denominator, 7, we have only to multiply both terms

2 X 3 by 3, which gives į

or, omitting to write the inter7 X 3

21' mediate work, we have

7 21 NOTE. - The same result might have been obtained thus :

21 1 21 3 1= hence of 1 must equal of

of is 7

and two times 7

21 3

; but, in practice, the first form will usually be found most

21 convenient. 7. Reduce & to fifteenths.

6

21

2

21'

21'7

21'

6

21

15 Solution. — Observing that the proposed denominator, 15, is of the

15 given denominator, 7, we have only to multiply both terms by

7' 2 2 X 27 44 llence,

44. 7 7 X 21 15' 7 15

or 2

2 or

8. Reduce š to halves.
Solution. — Observing that the proposed denominator, 2, is

of the

QI

5

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9

given denominator, 9, we have only to multiply both terms by

5 X

15 13 Hence,

9 X Reduce 9. to ninths.

14. $ to thirtieths. Ta to fourths.

15. 7 to twenty-sevenths. 11. Sto forty-fifths. 16. Í to ninths. 12. ☆ to twelfths.

17. i to sixty-fifths. 13. to forty-ninths. 18. 11 to twenty-seconds.

10.

151. Reduction to a Common Denominator. (a.) Fractions having their denominators alike are said to have a COMMON DENOMINATOR.

Thus, 4, §, 6, and have a common denominator, but $ and have not.

(6.) In reducing fractions having different denominators to a common denominator, (i. e., to equivalent ones having the same denominator,) we first select a convenient number for the common denominator, and then make the reductions as in the last article. (c.) As far as the denominator is concern

erned, one number may as well be selected for a common denominator as anothers but unless the number selected is a common multiple of all the given denominators, one or more of the resulting numerators will be likely to contain a fraction.*

(d.) To avoid such an inconvenience, and at the same time to avoid as far as possible the use of large numbers, it will usually be best to select the least common multiple of the given denominators for a common denominator.

1. Reduce ], 1, , and je to a common denominator.

* For it will be the product of a whole number multiplied by a frie: ticnal quantity.

nator.

Solution. - We select 72 for the least common denominator, because it is the least common multiple of the given denominators.

Then, since 72 = 9 times 8, we multiply both terms of the fraction by 9, which gives š= 43. Since 72 = 6 times 12, we multiply both terms of the fraction is by 6, which gives, &c.

Hence, š= **;=0; g = 13; and i=49.

(e.) Many adopt it as a general rule, to select the product of the given denominators for a common denominator; but it usually involves larger numbers than the preceding method, and is hence much less convenient. The following illustrates it.

Solution to preceding Example. — The product of 8, 9, 12, and 24, the given denominators, is 20736, which we select for the common denomi

To get this, we multiplied 8, the denominator of ], by 9 X 12 X 24, and therefore we multiply the numerator, 7, by the same numbers, which gives } = 18744. To obtain 20736, we multiplied 12, the denominator of P, by 8 X 9 X 24, the product of the other denominators, and therefore we multiply the numerator, 5, by the same numbers, which gives in 2436. To obtain 20736, we multiplied 9, the denominator of $, by, &c.

(f.) When any of the fractions to be reduced are compound or complex, they must first be reduced to simple ones, and the simple fractions should be reduced to their lowest terms, except when to do it would increase the labor of reducing to a common denominator.

(g.) Reduce the fractions in each of the following examples to a common denominator :

4 2 5 7
2.

and
5 3 6 15

3 5 1 1 4
3.

and
7' 9' 2' 6' 21

19 23 41 37 49
4.

and
24' 36' 72' 108

216
24 7 14 37
5.

and
25' 9' 15' 225 75

1 1 1 1 1 1 6.

and 2 3 4 5 6 12

44

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8.

25°

9.

10.

52

13 13 13

13

and
14' 21' 28'
2 3 5 8

12
of of and of
3 46 15' 9

16 314 7

7
of 71, and
8711
31 5 83 7

and
4 698 9
4 15 8 54 84 16

and
5

16' 32' 16777 32
7 12 13 8

of
21 44

and

164 il' 16' 22' ģ 22

100

11.

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12.

of

13.

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152. Addition and Subtraction of Fractions. (a.) Fractions, like other numbers, must be of the same denomination, in order to be added or subtracted.

(6.) To be of the same denomination, they must (121, e.) be fractions of the same unit, and also have a common denominator.

+ cannot be added in their present form, any more than can 3 pounds and 4 ounces.

of a yard and of an inch cannot be added in their present form, any more than can 3 yards and 3 inches.

(c.) When abstract fractions are given, (i. e., fractions of the abstract unit, as , 5.) they may, if simple, be reduced at once to a common denominator; but compound and complex fractions must be reduced to simple ones, and fractional parts of denominate numbers, as of a yard, s of a foot, must be reduced to fractions of the same unit, before they are reduced to a common denominator.

1. What is the sum of ++++? Solution. — By reducing to a common denominator, we have –

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24

21438

13 = 27

214

23 =

21.+ 20 + 16 + 22 + 18

424 2. What is the sum of 874 + 13+ 27+ 371 + 4872 ?

Solution. — The sum of the whole numbers is 212. Reducing the fractions to a common denominator, we have 1 + 8 + f + Io + 12 - 18 + 8 + 16 + 18+36=23, which added to 212 =

The following arrangement of the numbers shows the resemblance of the work to compound addition :

Ones. 36ths. 871 = 87 18 132

30 27

16 371 37 10 4872 48 21

21438 3. What is the sum of iz of a tont of a quarter ?

Solution. — By reducing the Iz of a ton to cwt. and qrs., and adding the & of a quarter, we should have the following written work:iz T. +qr.=l1 cwt. 23 qr. + & qr.

11 cwt. 33 qr.

=l1 cwt. qr.

12 lb. 8 oz. 4. What is the value of 343; -13821 ?

Solution. — By reducing the fractions to a common denominator, we have 3437 — 13831 = 3437. — 13892 = 20443, Or the work may be written thus :

Ones. 72ds.
3433 = 343 16 = Min.
1383 = 138 69

= Sub.
19 = 20443

= Rem. (d.) When several fractions are to be added, it will oftentimes be a saving of labor to consider at first only two of them, and then a third with the sum of these two, and then a fourth with the sum of these, and so on, till all are added. Care should be taken to couple them in such a way as to make the reductions easy.

Thus, in the 24th example below, & +f=1; 13 + 1 = 9; and 2 + i=2= Ans.

3

204

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