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11. It is required to find two numbers, whose difference is 25, and such that if the larger be multiplied by 7, and the smaller by 5, the difference of their products shall be 215?
XXV. Explanation of Negative Exponents.
It was observed above, that when the dividend and the divisor were different powers of the same letter, division is performed by subtracting the exponent of the divisor from that of the dividend : thus
Now 9 = 1. By the above principle" =qh = ao; there
sore a = l.
==; =61-) = 6° = 1; Q3
6 10 = 101- =10°=1;
a + b
= (a + b)". 10
= (a + b)° = 1* That is, any quantity having zero for its exponent, is equal to 1.
Hence it appears that a' has the same value as
The quantities a, a', a', a', a @m, a-, &c. have the same
1 1 1 value as as, a', a', 1,
* Exponents may be used for compound quantities as well as for simple; and multiplication and division may be performed on those which are similar, by adding and subtracting the exponents.
On this principle the denominator of a fraction, or any factor of the denominator may be written in the numerator by giving its exponent the sign This mode of notation is often very convenient; I shall therefore give a few examples of its application. 2 a
= 2 a b-c-. bc
By the principle explained above, 2 ab- x 6 c=2 ab-1+36-3+1 = 2 a b c =
2 a 68
2. Multiply 3 ac- -> by 3 a 6 d. 3. Multiply 5 a'c- by 2 ac'.
13 bd 4. Multiply
by 3 a c. 2 a ca
5. Multiply 2 a (b + d) by 3 a (b + d)".
3 a b 6. Multiply
by 8c* (2a-6d)" 4c(2a —bd)
3 a 7. Divide
by c. ca
Or thus, to divide 3 a c - by c*, is the same as to multiply it 1
or c, which gives the same result.
3 a c - d - X 4ec-3d-'= 12 a c-d-e=
36 10. Divide
2 b c
In this example the exponents to be subtracted had the sign, which in subtracting was changed to +. 11. Multiply 3 a(bc-d)" by
3(bc-d) 4 a 63
bco (6-2c)'. 12. Multiply
by 3c (6 — 2 c)
12 a 63 15 a' (bc-2)
> 13. Divide
by 16 62 ca
24 b' c3 14. Divide a°(176 + 3 d)?
5 a' (17b + 3 d) (a - b)
4(a - b)
XXVI. Examination of General Formulas.
When a question has been resolved generally, that is, by representing the known quantities by letters, we sometimes propose to determine what values the unknown quantities will take, for particular suppositions made upon the known quantities.
The two following questions offer nearly all the circunistances that can ever occur in equations of the first degree.
Two couriers set out at the same time from the points A and B, distant from each other a number m of miles, and travel towards each other until they meet. The courier who sets out from the point A, travels at the rate of a miles per hour; the other travels at the rate of 6 miles per
hour. At what distance from the points A and B will they meet?
Suppose C to be the point, and
x+y=AB=m Since the first courier travels x miles, at the rate of a miles per hour, he will be hours upon the road. The second cou
rier will be hours upon the road. But they travel equal times; therefore,
= 1/2 *=
abm т = x
b(a + b) Since neither of the quantities in these values of x and y has the sign-, it is impossible for either value to become nega
- az 뽕
tive. Therefore whatever numbers may be put in place of a, b, and m, they will give an answer according to the conditions of the question. In fact, since they travel towards each other, whatever be the distance of the places, and at whatever rate they travel, they must necessarily meet.
Suppose now that the two couriers setting out from the points A and B situated as before, both travel in the same direction towards D, at the same rates as before. At what distances from the points A and B will the place of their meeting, C, be?
Let x = the distance from A to C, and y =
B to C.
y=AC-BC=AB=m. The second equation expressing only the equality of the time will not be altered.
bm y =
6 m r=* 3
.b b(a-0) a- -6 Here the values of x and y will not be positive unless a is greater than b; that is, unless the courier, that sets out from A, travels faster than the other.
Suppose a = 8 and b = 4.