60. A cask which held 44 gallons was filled with a mixture of brandy, wine, and water. There were 10 gallons more than one half as much wine as brandy, and as much water as brandy and wine. How many gallons were there of each? 61. Two persons, A and B, travelling each with $80, meet with robbers who take from A $5 more than twice what they take from B; then B finds he has $26 more than twice what A has. How much is taken from each? Ans. From A, $69; from B, $32. 62. Four persons, A, B, C, and D, entered into partnership with a capital of $84816; of which B put in twice as much as A, C as much as A and B, and D as much as A, B, and C. How much did each put in? 63. In three cities, A, B, and C, 1188 soldiers are to be raised. The number of enrolled men in A is to that in B as 3:5; and the number in B to that in C as 8: 7. How many soldiers ought each city to furnish? Ans. A, 288; B, 480; C, 420. 64. Divide $65 among five boys, so that the fourth may have $2 more than the fifth and $3 less than the third, and the second $4 more than the third and $5 less than the first. 65. A merchant bought two pieces of cloth, one at the rate of $5 for 7 yards, and the other $2 for 3 yards; the second piece contained as many times 3 yards as the first times 4 yards. He sold each piece at the rate of $6 for 7 yards, and gained $24 by the bargain. How many yards were there in each piece? Ans. First, 84; second, 63. 66. A drover had the same number of cows and sheep. Having sold 17 cows and one third of his sheep, he finds he has three and a half times as many sheep as cows left. How many of each did he have at first? 67 A flour dealer sold one fourth of all the flour he had and one fourth of a barrel; afterward he sold one third of what he had left and one third of a barrel; and then one half of the remainder and one half of a barrel; and had 15 barrels left. How many had he at first? 68. A merchant bought a barrel of oil for $50; at the same rate per gallon as he paid, he sold to one man 15 gallons; then to another at the same rate two fifths of the remainder for $14. How many gallons did he buy in the barrel ? ど 69. Two pieces of cloth of the same length but different prices per yard were sold, one for $5 and the other for $7.50. If there had been 5 more yards in each, at the same rate per yard as before, they would have come to $15.474. How many yards were there in each? Ans. 21. 70. A and B began trade with equal sums of money. The first year A lost one third of his money, and B gained $750. The second year A doubled what he had at the end of the first year, and B lost $150, when the two had again an equal sum What did each have at Grst? 71. A man distributed among his laborers $2.50 apiece, and had $25 left. If he had given each $3 as long as his money lasted, three would have received nothing. How many laborers were there, and how much money did he have? Ans. 68 laborers, and $195. 72. A man who owned two horses bought a saddle for $35. When the saddle was put on one horse, their value together was double the value of the other horse; but when the saddle was put on the other horse, their value together was four fifths of the value of the first horse. What was the value of each horse? $ 175117 73. From a cask two thirds full 18 gallons were taken, when it was found to be five ninths full. How many gallons will the cask hold? = 162 74. A farmer had two flocks of sheep, and sold one flock for $60. Now a sheep of the flock sold was worth 4 of those left, and the whole value of those left was $8 more than the price of 8 sheep of those sold, and the flock left contained 40 sheep. How many sheep did the farmer sell, and what was the value of a sheep of each flock? Ans. Number sold, 15; value, $4 and $1. 75. A man has seven sons with 2 years between the ages of any two successive ones, and the sum of all their ages is ten times the age of the youngest. What is the age of each? 76. Divide 75 into two parts such that the greater in- 51 creased by 9 shall be to the less diminished by 4 as 3: 1. 77. Divide a into two parts such that the greater increased by b shall be to the less diminished by c as m: n. 78. What two numbers are as 34, while if 8 be added to each the sums will be as 5: 6? 79. Divide 127 into two parts, such that the difference between the greater and 130 shall be equal to five times the difference between the less and 63. 77 SECTION XIV. EQUATIONS OF THE FIRST DEGREE CONTAINING TWO UNKNOWN QUANTITIES. 109. INDEPENDENT EQUATIONS are such as cannot be derived from one another, or reduced to the same form. x 2 y Thus, x+y=10, += 5, and 4x + 3y = 40 — y are not independent equations, since any one of the three can be derived from any other one; or they can all be reduced to the form x+y= = 10. But x + y 4xy are independent equations. = 10 and 110. To find the value of several unknown quantities, there must be as many independent equations in which the unknown quantities occur as there are unknown quantities. From the equation x + y = 10 we cannot determine the value of either x or y in known terms. If y is transposed, we have X= 10y; but since y is unknown, we have not determined the value of x. We may suppose y equal to any number whatever, and then would equal the remainder obtained by subtracting y from 10. It is only required by the equation that the sum of two numbers shall equal 10; but there is an infinite number of pairs of numbers whose sum is equal to 10. But if we have also the equation 4 x y, we may put this value of y in the first equation, 10, and obtain x + 4x 10, or x = 2; then 4x-8=y, and we have the value of each of the unknown quantities. x + y = = ELIMINATION. 111. ELIMINATION is the method of deriving from the given equations a new equation, or equations, containing one (or more) less unknown quantity. The unknown quantity thus excluded is said to be eliminated. Transposing 4 x in (1) and dividing by 5, we have (3), which gives an expression for the value of y. Substituting this value of y in (2), we have (4), which contains but one unknown quantity; i. e. y has been eliminated. Reducing (4) ́we obtain (6), or Substituting this value of x in (3), we obtain (7), or x = 2. 3. Hence, RULE. Find an expression for the value of one of the unknown quantities in one of the equations, and substitute this value for the same unknown quantity in the other equation. NOTE. After eliminating, the resulting equation is reduced by the rule in Art. 102. The value of the unknown quantity thus found must be substituted in one of the equations containing the two unknown quantities, and this reduced by the rule in Art. 102. Find the values of x and y in the following equations: |