196 1 16 158 ON PROBLEMS RESULTING, &c. 34. Cloth, being wetted, shrinks up in its length and 8 in its width. If the surface of a piece of cloth is di 3 minished by 5 square yards, and the length of the 4 sides by 4 yards, what was the length and width of the cloth? 35. A certain number, less than 50, consists of two digits whose difference is 4. If the digits be inverted, the difference between the squares of the number thus formed and of the original number is 3960. Find the number. 11x + 40, 11x + 4 36. A plantation in rows consists of 10000 trees. If there had been 20 less rows there would have been 25 more trees in How many rows are there? a row. 37. A colonel wished to form a solid square of his men. The first time he had 39 men over: the second time he increased the side of the square by one man, and then he found that he wanted 50 men to complete it. How many men were there in the regiment? x2+39 = (x + 1)2 -50 A, not 3 XXII. INDETERMINATE EQUATIONS. 261. WHEN the number of unknown symbols exceeds that of the independent equations the number of simultaneous values of the symbols will be indefinite. We propose to explain in this Chapter how a certain number of these values may be found in the case of Simultaneous Equations involving two unknown quantities. Ex. To find the integral values of x and y which will satisfy the equation Here 3x + 7y=10. 3x=10-7y; and x=3-2y+m=3−2+6m+m=1+7m; or the general solution of the equation in whole numbers is x=1+7m and y=1-3m,. where m may be 0, 1, 2...... or any integer, positive or negative. and so on; from which it appears that the only positive integral values of x and y which satisfy the equation are 1 and 1. 262. It is next to be observed that it is desirable to divide both sides of the equation by the smaller of the two coefficients of the unknown symbols. Ex. To find integral solutions of the equation Hence Now if if if x=n-2m=n−2+4n=5n-2; y=6−x+m=6-5n+2+1−2n=9-7n. n=0, x=-2, y = 9; n = 1, x= 3, y= 2; and so on. 263. In how many ways can a person pay a bill of £13 with crowns and guineas? Let x and y denote the number of crowns and guineas. and higher values of m will give negative values of x. Thus the number of ways is three. 264. To find a number which when divided by 7 and 5 will give remainders 2 and 3 respectively. 13. Find two fractions with denominators 7 and 9 and 14. Find two fractions with denominators 11 and 13 and 15. In how many ways can a debt of £1. 9s. be paid in florins and half-crowns ? 16. In how many ways can £20 be paid in half-guineas and half-crowns? 17. What number divided by 5 gives a remainder 2 and by 9 a remainder 3? 18. In how many different ways may £11. 15s. be paid in guineas and crowns? 19. In how many different ways may £4. 11s. 6d. be paid with half-guineas and half-crowns? 20. Shew that 323x-527y = 1000 cannot be satisfied by integral values of x and y. 21. A farmer buys oxen, sheep and hens. The whole number bought was 100, and the whole price £100. If the oxen cost £5, the sheep £1, and the hens 1s. each, how many of each had he? Of how many solutions does this Problem admit? 22. A owes B 4s. 10d.; if A has only sixpences in his pocket and B only fourpenny pieces, how can they best settle the matter? 23. A person has £12. 4s. in half-crowns, florins and shillings, and the numbers of coins of each kind are respectively as 7, 5, 3. Find the number of coins of each kind. 24. In how many ways can the sum of £5 be paid in exactly 50 coins, consisting of half-crowns, florins and fourpenny pieces? 25. A owes B a shilling. A has only sovereigns and B has only dollars worth 4s. 3d. each. How can A most easily pay B? 26. Divide 25 into two parts such that one of them is divisible by 2 and the other by 3. X27. In how many ways can I pay a debt of £2. 9s. with crowns and florins? . 28. Divide 100 into two parts such that one is a multiple of 7 and the other of 11. 29. The sum of two numbers is 100. The first divided by 5 gives 2 as a remainder, and if we divide the second by 7 the remainder is 4. Find the numbers. 30. Find a number less than 400, which is a multiple of 7, and which when divided by 2, 3, 4, 5, 6, gives as a remainder in each case 1. |
