NOTES. 1. If we restrict the enunciation of the problem to its arithmetical sense, in which "added" means numerical increase," the first value of x only will satisfy the conditions of the problem. 2. If we give to "added" its algebraical signification (when it may mean subtraction as well as addition), the problem may be thus stated: -To find a number such that twice its square diminished by three times the number shall give 65. The second value of x will satisfy this enunciation; for 3. The root which results from giving the plus sign to the radical is generally an answer to the question in its arithmetical sense. The second root generally satisfies the problem under a modified statement. Thus, in the example it was required to find a number of which twice the square added to three times the number shall give 65. Now, in the arithmetical sense, "added" means increased; but in the algebraic sense it implies diminution when the quantity added is negative. In this sense, the second root satisfies the enunciation. 5. A certain person purchased a number of yards of cloth for 240 cents. If he had purchased 3 yards less of the same cloth for the same sum, it would have cost him 4 cents more per yard. How many yards did he buy? -- If for 240 cents he had purchased three yards less, that is, x 3 yards, the price per yard, under this hypothesis, would have been But by the conditions this last cost must exceed the denoted by 240 X- 3 first by 4 cents. Therefore we have the equation NOTES. - 1. The value x' = 15 satisfies the enunciation in its arith metical sense; for if 15 yards cost 240 cents, 240 ÷ 15 16 cents, the price of one yard; and 240 ÷ 12 = 20 cents, the price of 1 yard under the second supposition. 2. The second value of x is an answer to the following problem: : A certain person purchased a number of yards of cloth for 240 cents. If he had paid the same for three yards more, it would have cost him 4 cents less per yard. How many yards did he buy ? the same equation as found before. Hence A single equation will often state two or more arithmetical problems. This arises from the fact that the language of algebra is more comprehensive than that of arithmetic. 6. A man, having bought a horse, sold it for $24. At the sale he lost as much per cent on the price of the horse as the horse cost him dollars. What did he pay for the horse? Let x= x-24 = the number of dollars that he paid for the horse. Then the loss he sustained. But as he lost x per cent by the sale, he must have lost upon each dollar, and upon x dollars he lost a 100 Both of these roots will satisfy the problem: for, if the man gave $60 for the horse, and sold him for $24, he lost $36. From the enunciation he should have lost 60 per cent of $60; that is, Had he paid $40 for the horse, he would have lost by the sale $16. From the enunciation he should have lost 40 per cent of $40; that is, 7. The sum of two numbers is 11, and the sum of their squares is 61. What are the numbers? Ans. 5 and 6. 8. The difference of two numbers is 3, and the sum of their squares is 89. What are the numbers? Ans. 5 and 8. 9. A grazier bought as many sheep as cost him £60; and, after reserving fifteen out of the number, he sold the remainder for £54, and gained 2s. a head on those he sold. How many did he buy? Ans. 75. 10. A merchant bought cloth, for which he paid £33 15s., which he sold again at £2 8s. per piece, and gained by the bargain as much as one piece cost him. How many pieces did he buy? Ans. 15. 11. The difference of two numbers is 9, and their sum multiplied by the greater is equal to 266. What are the Ans. 14 and 5. numbers? 12. Find a number such that, if you subtract it from 10 and multiply the remainder by the number itself, the product will be 21. Ans. 7 or 3. 13. A person traveled 105 miles. If he had traveled 2 miles an hour slower, he would have been 6 hours longer in completing the same distance. How many miles did he travel per hour? Ans. 7 miles. 14. A person purchased a number of sheep, for which he paid $224. Had he paid for each twice as much, plus $2, the number bought would have been represented by twice what was paid for each. How many sheep were purchased? 15. The difference of two numbers is 7, multiplied by the greater is equal to 130. numbers? Ans. 32. and their sum What are the Ans. 10 and 3. 16. Divide 100 into two such parts that the sum of their squares shall be 5392. Ans. 64 and 36. 17. Two square courts are paved with stones a foot square. The larger court is 12 feet larger than the smaller one, and the number of stones in both pavements is 2120. How long is the smaller pavement? Ans. 26 feet. 18. Two hundred and forty dollars are equally distributed among a certain number of persons. The same sum is again distributed amongst a number greater by 4. In the latter case each receives ten dollars less than in the former. many persons were there in each case? How Ans. 8 and 12. 19. Two partners, A and B, gained $360. A's money was in trade 12 months, and he received for principal and profit $520. B's money was $600, and was in trade 16 months. How much capital had A? Ans. $400. EQUATIONS CONTAINING MORE THAN ONE UNKNOWN QUANTITY. 173. Two simultaneous equations, each of the second degree, and containing two unknown quantities, will, when combined, generally give rise to an equation of the fourth degree. Hence only particular cases of such equations can be solved by the methods already given. |