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.. X =

Substituting this value in the original equation, we obtain

7x + 24 – 84m=220;
=28+12m ..

(B). Equation (A) shows that m may be 0 or have any negative ral value but cannot have a positive integral value.

Equation (B) shows in addition that m may be O but cannot have a negative integral value greater than 2. Thus the only positive integral values of x and y are obtained by placing m=0, -1, -2. The complete solution may be exhibited as follows:

m= 0, -1, -2,
x=28, 16, 4,

y= 2, 9, 16. Example 2. Solve 5x — 14y=11 in positive integers .....(1). Proceeding as in Example 1, we obtain

4y+1 X=2+2y+

5

4y+1
... 3—27-2=1 integer.

5
Now multiplying the numerator by 4, we obtain

16y +4

integer;

5 that is,

y+4
3y + integer.

5 Let

=m, an integer; 5

.: y= 5m-47 and from (1)

x=14m-9S This is called the general solution of the equation, and by giving to m any positive integral value, we obtain an unlimited number of values for x and y: thus we have

m=1, 2, 3, 4.....
y=1, 6, 11, 16.

@=5, 19, 33, 47. From Examples 1 and 2 the student will see that there is a further limitation to the number of solutions according as the terms of the original equations are connected by + or If we have two equations involving three unknown quantities, we can easily combine them so as to eliminate one of the unknown quantities, and can then proceed as above.

=

y+4

INDETERMINATE EQUATIONS OF THE FIRST DEGREE. 215

EXAMPLES XXX. Solve in positive integers : 1. 3x+8y=103. 2. 5x+2y=53. 3. 73+12y=152. 4. 13x+11y=414. 5. 23x+25y=915. 6. 41x+47y=2191.

Find the general solution in positive integers, and the least values of x and y which satisfy the equations :

7. 5x – 7y=3. 8. 6x — 13y=1. 9. 8x— 21y=33. 10. 17y - 13x=0. 11. 19y – 23x=7. 12. 777-30x=295.

13. A farmer spends $752 in buying horses and cows; if each horse costs $37 and each cow $23, how many of each does he buy?

14. In how many ways can $100 be paid in dollars and halfdollars, including zero solutions ?

15. Divide 81 into two parts so that one may be a multiple of 8 and the other of 5.

16. Find a number which being divided by 39 gives a remainder 16, and by 56 a remainder 27. How many such numbers are there?

17. Divide 136 into two parts one of which when divided by 5 leaves remainder 2, and the other divided by 8 leaves remainder 3.

18. I buy 40 animals consisting of rams at $4, pigs at $2, and oxen at $17: if I spend $301, how many of each do I buy ?

CHAPTER XXXI.

THE THEORY OF INDICES.

231, HITHERTO all the definitions and rules with regard to indices have been based upon the supposition that they were positive integers; for instance (1)

al4=a.a.a...... to fourteen factors.
(2) al4 x a=q14+3 = a17.
(3) al4 ;-a3=a14–3 =all.

(a14)3=a143=a42 The object of the present chapter is twofold : first, to give general proofs which shall establish the laws of combination in the case of all positive integral indices; secondly, to explain how, in strict accordance with these laws, intelligible meanings may be given to symbols whose indices are fractional, zero, or negative.

We shall begin by proving, directly from the definition of a positive integral index, three important propositions,

232. DEFINITION. When m is a positive integer, am stands for the product of m factors each equal to a.

233. PROP. I. To prove that am xa"=am+n, when m and n are positive integers. By definition, am=a.a. to m factors; an=

to n factors; am xan=(a.a.a to m factors) (a.a.&... to n factors)

to mtn factors =am+n, by definition. Cor. If p is also a positive integer, then

am xan XaP=am+n+p; and so for any number of factors.

a

=a.a.a......

=a.a. a

am

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a. a. a

234. PROP. II. To prove that am -- a"=a"-", when m and n are positive integers, and m>n.

a.a.a... to m factors
am • a^=

to n factors
=a.a. a ... to m n factors

=am-n. 235. Prop. III. To prove that (am)"=amn, when mand n are positive integers. (am)"=am.am. to n factors

=(a.a.a ... to m factors)(a.d.d ... to m factors)...... the bracket being repeated n times,

to mn factors =amn.

am

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236. These are the fundamental laws of combination of indices, and they are proved directly from a definition which is intelligible only on the supposition that the indices are positive and integral.

But it is found convenient to use fractional and negative indices, such as ab, a-?; or, more generally, a?, a-n; and these have at present no intelligible meaning. For it is plain that the definition of am, [Art. 232] upon which we based the three propositions just proved, is no longer applicable when mis fractional, or negative.

Now it is important that all indices, whether positive or negative, integral or fractional, should be governed by the same laws. We therefore determine meanings for symbols such as a', a-n, in the following way: we assume that they conform to the fundamental law, am xan=am+n, and accept the meaning to which this assumption leads us. It will be found that the symbols so interpreted will also obey the other laws enunciated in Props. II. and III.

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237. To find a meaning for a?, p and q being positive integers.

Since am xan=am+n is to be true for all values of m and n, by replacing each of the indices m and n by ??

we have

a' ??

2p a? x al=a?! =al.

P

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P

P

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P P P 2p P

8p Similarly, al x al xar=al xal=a! =a. Proceeding in this way for 4, 5, q factors, we have

al xa x al to q factors=ae; that is,

(ar)e=a.

=”. Therefore, by taking the qth root,

[blocks in formation]

P

or, in words, al is equal to “the qth root of a.

Examples. (1) 27=25.

5

1

3

2

5

2 5

3

a

a

(2) a3= a.
(3) 42=143=164=8.
(4) až x að=až+6=až.
2

3a +
(5) k2 x k3= k2*3= k 3
2 1 15

2 1
(6) 323 32 x 425 76 = 12a3+ 12+

=122713 238. To find a meaning for ao.

Since am xa”=am+n is to be true for all values of m and n, by replacing the index m by 0, we have

a'xa"=Qo+n

-a",

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-1. Hence any quantity with zero index is equivalent to 1.

[See Art. 47.] Example. b-exxe

P=29=1. 239. To find a meaning for a-".

Since am xa" =am+n is to be true for all values of mand n, by replacing the lex in by - n, we have

a-n xa"=a-ntno

=QO
= 1.

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