his son; and that both of their ages put together, would amount to 49 years. What was the age of each? Ans. Son's age 7 years; father's 42. 5. A farmer said that he had four times as many cows as horses, and five times as many sheep as cows; and that the number of all of them was 100. How many had he of each sort. Ans. 4 horses; 16 cows; and 80 sheep. 6. A boy told his sister that he had ten times as many chestnuts as apples, and six times as many walnuts as chestnuts. How many had he of each sort, supposing there were 639 in all. Ans. 9 apples; 90 chestnuts; and 540 walnuts. 7. A school girl said that she had 120 pins and needles; and that she had 7 times as many pins as needles. How many had she of each sort? Ans. 15 needles, and 105 pins. 8. A teacher said that her school consisted of 64 scholars; and that there were three times as many in arithmetic as in algebra, and four times as many in grammar as in arithmetic. How many were there in each study? Ans. 4 in algebra; 12 in arithmetic; and 48 in grammar. 9. Two men, who are 560 miles apart, start to meet each other. One goes 30, and the other goes 40 miles a day. In how many days will they meet? Ans. 8 days. Remarks. Each will travel a days. The first will go x times 30 miles, and the second will go x times 40 miles; and both together will go the whole distance. It is also evident that x times 30 is the same as 30 times x; &c. 10. A teacher had four arithmeticians who performed 80 sums in a day. The second did as many as the first, the third twice as many, and the fourth as much as all the other three. How many did each perform? Ans. The first and second, each 10; the third, 20; and the fourth, 40 11. A person said that he was $450 in debt. That he owed A a certain sum, B twice as much, and C twice as much as to A and B. How much did he owe each? Ans. To A $50, to B $100; and to C $300. 12. A person said that he was owing to A a certain sum; to B four times as much; and to C eight times as much; and to D six times as much; so that $570 dollars would make him even with the world. What was his debt to A? Ans. $30. For 13. A man bought 3 sheep and 2 cows for $60. each cow, he' gave 6 times as much as for a sheep. How much did he give for each? Remarks. If x= price of a sheep, all the sheep will cost three times as much, or 3x. In the same manner One cow both cows will cost twice as much as one cow. will cost 6x, and 2 cows will cost 12x. Ans. $4, price of a sheep; and $24 price of a cow. 14. A gentleman hired 3 men and 2 boys for one day. He gave five times as much to a man as he gave to a boy; and for all of them he gave $6.80. What was the wages of each? Ans. A boy's wages was 40 cents, and a man's wages, $2. 15. A boy bought some oranges and some lemons for 54 cents. There was an equal number of each sort, but the price of an orange was twice the price of a lemon. How much money did he spend for each sort? Ans. 18 cents for lemons; and 36 cents for oranges. 16. A boy bought some apples, some pears, and some peaches, an equal number of each sort, for 72 cents. The price of a pear was twice that of an apple, and the price of a peach was 3 times that of an apple. How much money did he give for each kind? Ans. 12 cents for apples; 24 cents for pears; 36 cents for peaches. 17. A farmer hired three labourers for $50.00; giving to the first $2.00 a-day, to the second $1.50, and to the third $1.00. The second worked three times as many days as the first; and the third twice as many days as the second. How many days did each work? Ans. The first, 4; second, 12; and third, 24 days. i 18. A gentleman bought some tea, coffee, and sugar, for $7.04; giving twice as much a pound for coffee as for sugar, and five times as much for tea as for coffee; and there were 20 pounds of sugar, 12 pounds of coffee, and 2 pounds of tea. What was the price of each? Ans. 11 cents for sugar; 22 cents for coffee; and 110 cents for tea. ADDITION OF COMPOUND QUANTITIES. §42. When two or more expressions that consist of several terms, are to be added together, the operation is represented by connecting them with one another by means of the sign. Thus, a―x is added to y+7 in the following manner: a-x+y+7, or y+7+a-x. For to a-x we add first y, and then 7; or to y+7 we add a, and then because we ought to have added a-x, we see that we have added a too much, and therefore subtract it. §43. The above example shows that it is of no consequence in what order we write the terms. Their place may be changed at pleasure, provided their signs be preserved. §44. After compound quantities have been added, their terms may be united according to the rule §28. EXAMPLES. 1. Add the following compound quantities. 2a-8x, x-3a, -4a-2x, 4x-a. Ans. 6a-5x. 2. Add 2-x+4y, 3+3x-y, -30-x-2y, and 1-2x+3y-10% together. Ans. 24-x+4y-10%. 3. Add 3x+5y-6%, -2x-8y-9z, 20+2y-3z and x-y+z-4 together. Ans. 22x-2y-17-4. 4. Add 3-2y+z, 4y-2z+5, 2-z-y, and 2z-y -10. Ans. Nothing. 5. Add 7x-64-5z-8-g 3-g-3y-x 3z-1-g+3y-2x x+8y-5x+9+g. Ans. 4x+5g+3y+2. 6. Add 3b-a-c-115d+6e-5y 3a+27e-d-3c-2b 3e-7y—8c+5b 17c-6b-7a+9d-5e+11y -2d-6e-5c-9y-3a+x. Ans. 37e-8a-109d-10y+x. 7. Add 4-3a+x-7+4y 8+6y-3x+6a+32 7a+3y+4a+12-x 8. Add a-x+y+r-14-n 3n-y+2+6a-2+3x+r 62+7r-a-y-2n 3y+2a+1+6r+4n x−7+x+y+4-n+r. Ans. 42 TRANSPOSITION. 1. BY SUBTRACTION. §45. It often happens that in the first member of the equation, some number has been added to the x's in order to make them equal to the last member. equation, x+16=46, we see that 16 has been added to x, Thus, in the to make it equal to 46. §46. Now if x with 16 added, is equal to 46; then x alone must be 16 less than 46; that is, 46-16. So, that if we find, that x+16=46, we may know that x=46—16; or what is the same, x=30. $47. But this may be proved another way. It is very plain that if we subtract a quantity from one member of an equation, and then subtract the same quantity from the other member of the equation; it will still be the fact that the two members are equal to one another. Thus, a half dollar 50 cents. Subtract 2 cents from each member. Then a half dollar 2 cents 50 cents each of them is equal to 48 cents. 2 cents; for $48. Now, with the equation that we had above, +16=46. Subtracting 16 from both members, x+16-16=46-16. Now, in the first member of the equation, we have+16 -16, which is of no value at all, for +16 and -16 balance each other as has been seen in Ex. 1 under §28. Therefore the equation is reduced to Uniting terms in the last member, EQUATIONS.-SECTION 2. x=46-16. x=30. 1. Suppose 2+8+3x-56; what is the value of x? Uniting terms, 4x+8=56. |