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XX. Sometimes division may actually be performed when both divisor and dividend are compound quantities. Since division is the reverse of multiplication, the proper method to discover how to perform it, is to observe how a product is formed by multiplication.

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8 a” b”— 12 a'b' c + 4 a” b” co-H4 a'b' c-6 a” b” co-H 2 a b'c'.

Observe that each term of the multiplier is multiplied separately into each term of the multiplicand. The product therefore must consist of a number of terms equal to the product of the number of terms in the multiplicand by the number of terms in the multiplier. If the product be divided by the multiplicand, the multiplier must be reproduced, and if by the multiplier, the multiplicand must be reproduced.

The three terms 8 a' bo — 12 o'b' c +4 a'b'c' of the product were produced by multiplying the three terms of the multiplicand by the first term of the multiplier, 4 a” b”. Therefore, if these three terms be divided by 4 a bo, the quotient will be the multiplicand.

Again, the three terms

4 a'b' c – 6 a” b” co-H 2 a b'c'

of the product were formed by multiplying each term of the multiplicand by 2 a b c. Therefore, if these three terms be divided by 2 ab c, the quotient will be the multiplicand.

Hence we see that the whole division might be performed by any one term of the divisor, if all the terms of the dividend which depend on that term and the quotient could be ascertained. This cannot often be done by inspection; for in many products, though at first there are as many terms as there are in the multiplicand and multiplier together, some of the terms are united together by addition or subtraction, and some disappear entirely. Even if all the terms did remain entire, they could not be easily distinguished.

However, one term may always be distinguished, and from it one term of the quotient may be obtained.

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First, it is evident that the highest power of either letter in the dividend, must have been produced by multiplying the highest power of that letter in the divisor by the highest power of the same letter in the quotient; for in order to produce the dividend, each term of the divisor must be multiplied by every term of the quotient. Therefore, if 4 at be divided by 2 a” it must give a term of the quotient. Or, if—bo be divided by b" it must give a term of the quotient. Let the quantities be arranged according to the powers of the letter a.

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I divide 4 a” by 2 a”, which gives 2 a” for the first term of the quotient. Now in forming the dividend, every term of the divisor was multiplied by this term of the quotient, therefore I multiply the divisor by this term, by which means l find all the terms of the dividend, which depend on this term. They are.

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Here is a term 6 a” b which is not in the dividend, this must have disappeared in the product. The term 2 a bo is not found alone, but it is like 9 a” b” and must have disappeared by uniting with some other term to form that. I subtract these three terms from the dividend, and there remains

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which does not depend at all on the term 2 a” of the quotient, but which was formed by multiplying each remaining term of the quotient by all the terms of the divisor. This then is a new dividend, and to find the next term of the quotient we must proceed exactly as before; that is, divide the term of the dividend containing the highest power of a, which is 6 a'b, by 2 a” of the divisor, because this must have been formed by multiplying 2 a” by the highest remaining power of a in the quotient. This gives for the quotient + 3 a b. I multiply each term of the divisor by this, and subtract the product as before, and for the same reason. The remainder is

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which depends only on the remaining part of the quotient. The highest power of a, viz. 2 a bo, must have been produced by multiplying some term of the quotient by 2 at of the divisor; therefore I divide by this again, and obtain—bo for the quotient. I multiply by this and subtract as before, and there is no remainder, which shows that the division is completed. By the above process I have been enabled to discover all the terms of the dividend produced by multiplying the first term of the divisor by each term of the quotient. If both be arranged according to the powers of the letter b, and the same course pursued, the same quotient will be obtained, but in a reversed order. In the division the term — 2 a bo has the sign —. Here we must observe that the divisor and quotient multiplied together must reproduce the dividend. If + a b be divided by + a, the quotient must be + b, betause + a x + b gives + a b.

If–ab be divided by + a, the quotient must be o, because + a X b gives—a b. If + a bibe divided by — a, the quotient must be — b, because a X b gives + ab. If– a b be divided by —a, the quotient must be + b, because — a X + b gives—a b. The rule for signs therefore is the same as in multiplication. When the signs are alike, that is, both + or both —, the sign of the product must be + ; but when the signs are unlike, that is, one + and the other —, the sign of the quotient must be —. By the reasoning above we derive the following rule for division of compound numbers. Arrange the dividend and divisor according to the powers of some letter. Divide the first term of the dividend by the first term of the divisor, and write the result in the quotient. o all the terms % the divisor by the term of the quotient thus found, and subtract the product from the dividend. The remainder will be a new dividend, and in order to find the next term of the quotient, proceed exactly as before; and so on until there is no remainder. Sometimes, however, there will be a remainder, such that the first term of the divisor, will not divide either term of it; in which case the division can be continued no farther, and the remainder must be written over the divisor in the form of a fraction, and annexed to the quotient as in arithmetic. Divide 2 a”— 11 at b + 11 a” b” + 13a*b* by 2 a-b.

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In this example,

the division may be continued until the re

mainder is 45, which cannot be divided by a, therefore it most

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b as a fraction and added to

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XXI. Equations.

The above rules are sufficient to solve all equations of the

first degree.

1. Find the value of r in the equation

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2 a -b.

b” as

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First, clear it of fractions by multiplying by the denomina

torS.

• Let the learner prove his results by multiplication.

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