65. A man bought a certain number of sheep for $290. Having lost five of them, he sold one quarter of the remainder for $ 63, making a profit of 5 per cent on the sale of these. How many

67. (a + b)x + by = ax + (a + b)y = a3 + b3.

70. A number of articles were bought for $25, and sold again at an advance of 12 cents in the price of each; and at the advanced price $25 was received for the sale of all but 10 of the articles. How many were there?

71. (3x+1)}(2 x − 3) = √(6x1)- (7x − 3).

73. √(x+1)+ √2 x = √(6x + 1).

74. x2xy + 3y = 11, y2 — xy − 3x + 1 = 0.

75. The value of 185 coins consisting of dollars, dimes, and half-dimes amounts to $30.50. If there were twice as many dimes, half as many half-dimes, and three times as many dollars, the total value of the coins would be $72. How many coins are there of each kind?

80. 1120 square feet of paper will just cover the four walls of a room which is 8 feet longer than it is wide; but, if the room were 4 feet higher, the same quantity of paper would just cover the two smaller and one of the larger walls. What are the dimensions of

83. (x − 5)2 + (y − 6)2 = 2(xy − 40), x = y + 1.

84. 4 xy 96 − x2y2, x + y = 6.

=

85. A merchant gained as many eagles on the sale of a certain quantity of coal as there were half-dollars in the cost price of a ton, or half-dimes in the retail price of a cwt. How many tons did he sell?

90. A rectangular enclosure is half an acre in area, and its perimeter is 198 yards. Find the length of its sides.

92. x+2y+3 z = 4, x + 3y + 2 = 4 z, x + 2 z + 3 = 4y.

93. (ax)+ √(x − b) = √(a — b).

94. (x-3)2+(y-3)2 = 34, xy-3(x+y)= 6.

95. Two trains start at the same instant, the one from B to A, the other from A to B, the distance between A and B being 100 miles. The trains meet in 1 hour 15 minutes, and one train gets to its destination 1 hour 20 minutes before the other. Find the rates of the trains.

100. A starts to walk from P to Q at 10 A.M., B starts from Q to walk to P at 10.24 A.M. They meet 6 miles from Q. B stops 1 hour at P, and A stops 2 hrs. 54 min. at Q, and returning they meet midway between P and Q at 6.54 P.M. Find the distance

from P to Q.

101. y2+z2 = 22 + x2 = x2 + y2 = axyz.

102. ax(y + z) = by (z + x) = cz(x + y) = xyz.

104. Resolve x3 + y3 + z3 − 3 xyz into the three factors

x + y + z, x + wy + w2z, x + w2y + wz,

and determine three alternative relations between a, b, and c, one of which must be satisfied in order that the equations

CHAPTER XX.

POWERS AND ROOTS.

195. The process by which the powers of quantities. are obtained is called involution; and the inverse process, by which the roots of quantities are obtained, is called evolution. [Art. 9.]

INVOLUTION.

196. When m and n are any positive integers, we have by definition

•*. am×a”=(aaaa... to m factors) × (aaaa... to n factors)

= aaaa... to m + n factors

=

: am+n,

by definition.

Hence when m and n are any positive integers,

am+a" = am+n.

Thus, the index of the product of any two powers of the same quantity is the sum of the indices of the factors. This result is called the Index Law.

From the Index Law we have

and so on, however many factors there may be.

Hence

am xan X ⱭP.

am+n+p+....

Thus, the index of the product of any number of powers of the same quantity is the sum of the indices of the factors.

197. We have

am/an

=

a xa xa xa... to m factors

a xa xa x a... to n factors

Now, if m is greater than n, the n factors of the denominator can be cancelled with n of the factors of the numerator: we then have m n factors left in the

numerator.

Thus, when m is greater than n,

am/an = am-n.

If, however, n is greater than m, the m factors of the numerator can be cancelled with m of the factors of the denominator: we then have n

m factors left in

198. To find (am)" when m and n are positive integers.

By definition

Hence

(am)n = am × am × am ו to n factors

= am+m+m+... to n terms

= amn

...

by Art. 196.

(am)n = amn.