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Here, the least common multiple of 4cx2, 2x, and Sac ; (Art. 147), is Sacra; then, Sacox2

X 3a2b=2ac +-3a2b=la’bc 4022 8ac2x2

new numeraXy=4aco Xy=4ac2y 2x

tors; Sac? x2

X 5?=x2 X 512=5 x 4 Saca

6a3bc Hence

and

are

the 8ac2x2 Sac??? fractions required.

4acʻy

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8aca

2

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5.0 a

1 Ex. 6. Reduce

and to a comat 3

2x' mon denominator. 30x2

- 2x 3 3a +32 Ans.

and Gax +-6x2 6ax+6x2 6ax +6x2

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Ex. 8. Reduce

3 mon denominator.

4.0 +4. Ans.

3.2 +6x +3

120+12?

120+12

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a

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2c2

3a-202 Ex. 9. Reduce

and ct

d' common denominator.

adx 2bc x Ans.

and bdx bdx

3abd bdx;

X2

a

54'

2025

40--15 Ex. 10. Reduce

and 7+

> 3x

2 to a common denominator.

6x9 30axy-10x2 y + 50y
Ans.
30xy

30xy 60xy-15xy

30xy

and

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a

36 Ex. 11. Reduce

5x

and a2 -22' 4a-4x'

at to other equivalent fractions having the least common denominator.

4a 3ab +3.0 Ans.

20ax-20x2

and 4a2-4x2' 442 4x2

40-4X2

1

1 Ex. 12. Reduce

and a2 + 2ax txa

ao-c2 : 5y

to the least common denominator. a4 — X4 Ans.

au taxa ax—203 a3 +-axtax+23 a5

-ax4 tao -25? as --- ax 4 t-aux

5ay + 5xy and

a5 -- ax 4 fax 20

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154. Reduce the fractions, if necessary, to a common denominator, by the rules in the last case, then add all the numerators together, and under their sum put the common denominator ; bring the resulting fraction to its lowest terms, and it will be the sum required.

2c 5x Ex. 1. Add

and 3' 7'

together.

9 2x X7X9=126x

126x +135x+212 2820 5% X 3X9=1350

189

139 X X 7 X3= 21x

932
+ is the sum required.

189 3X7X9189

4a

a
2u

56 Ex. 2. Add

and together. b' 36'

12a2b+8ab-+-1563 a X 36 X 4a=12a26

12ab2 2a X6 X 4a= 8a2b | 20a2b+1563

=(dividing by 6) 56 X 36 Xb= 1563 12ab2

20a? +1562

is the sum b X 36 X 4a=12ab? 12ab

quired.

le

Or, the least common multiple of the denominators may be found, and then proceed, as in (Art. 153).

It is generally understood that mixed quantities. are reduced to improper, fractions, before we perform any of the operations of Addition and Subtraction. But it is best to bring the fractional parts only to a common denominator, and to affix their sum or difference to the sum or difference of the integral part interposing the proper sign.

322 Ex. 3. It is required to find the sum of a

b: 2ax and b+

с
3х2
ab - 3x^

2ax bc + 2ax Here, a

and 6+ b

b Then, (ab-3x2) Xc=abc - 3cx 2

numerators. (bc+ 2ax) xb=b2c-+-2abx

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с

bc

Xe=bc= denominator.
abc - 3cx? +-62c-f-2abx abc-bac

+

bc 2abx - 3cx 2

2abx - 3022 =atbota bc

bc is the sum required.

Or, bringing the fractional parts only to a comnon denominator, Thus, 3.2"Xc= 3cxa

numerators,
2ax Xb=2abx
And bXc=bc common denominator.

Зcx2
2abx

2abx – 3cx Whence a +6+ =a+b+

bc

'bc the sum.

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Ex. 4. It is required to find the sum of 50+ -2

2x-3 and 4x 3

50 Here, (2-2) X 5x=5x2 -- 10x

numerators. (1-3) X3.=6 - 9

}

And 3 X 5x=15x common denominator. 5x2-100

6x --- 9 Whence 5x+

+4x

=9x + 1500

15.0 5x2 +10x 9462 5x2-16x +9 + 9x +

the sum 150 152

15 x required. 6 x 9

9-62 Here, is evidently

(Art. 128); 150

152 but we might change the fractions into other equivalent forms before we begin to add or subtract ; thus, the fractional part of the proposed quantity

2x - 3 4x

may be transformed by changing the

5x signs of the numerator, (Art. 128), and the quantity

3-20 itself can be written thus, 4x+ : It is well to

50 keep this transformation in mind, as it is often necessary to make use of it in performing several algebraical operations.

3a2 2a b Ex. 5. Add and together.

5

7

105a27-28ab+10b2 Ans.

706

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