EXAMPLES. XXVIII. 1. Divide the number 60 into two parts such that their product may be 864. 2. The sum of two numbers is 60, and the sum of their squares is 1872: find the numbers. 3. The difference of two numbers is 6, and their product is 720 find the numbers. 4. Find three numbers such that the second shall be two-thirds of the first, and the third half of the first; and that the sum of the squares of the numbers shall be 549. 5. The difference of two numbers is 2, and the sum of their squares is 244: find the numbers. 6. Divide the number 10 into two parts such that their product added to the sum of their squares may make 76. 7. Find the number which added to its square root will make 210. 8. One number is 16 times another; and the product of the numbers is 144: find the numbers. 9. One hundred and ten bushels of coals were divided among a certain number of poor persons; if each person had received one bushel more he would have received as many bushels as there were persons: find the number of persons. 10. A company dining together at an inn find their bill amounts to £8. 15s.; two of them were not allowed to pay, and the rest found that their shares amounted to 10 shillings a man more than if all had paid: find the number of men in the company. 11. A cistern can be supplied with water by two pipes; by one of them it would be filled 6 hours sooner than by the other, and by both together in 4 hours: find the time in which each pipe alone would fill it. 12. A person bought a certain number of pieces of cloth for £33. 15s., which he sold again at £2. 8s. per piece, and he gained as much in the whole as a single piece cost: find the number of pieces of cloth. 13. A and B together can perform a piece of work in 14 days; and A alone can perform it in 12 days less than B alone: find the time in which A alone can perform it. 14. A man bought a certain quantity of meat for 18 shillings. If meat were to rise in price one penny per lb., he would get 3lbs. less for the same sum. Find how much meat he bought. 15. The price of one kind of sugar per stone of 14 lbs. is 18. 9d. more than that of another kind; and 8 lbs. less of the first kind can be got for £1 than of the second: find the price of each kind per stone. 16. A person spent a certain sum of money in goods, which he sold again for £24, and gained as much per cent. as the goods cost him: find what the goods cost. 17. The side of a square is 110 inches long: find the length and breadth of a rectangle which shall have its perimeter 4 inches longer than that of the square, and its area 4 square inches less than that of the square. 18. Find the price of eggs per dozen, when two less in a shilling's worth raises the price one penny per dozen. 19. Two messengers A and B were despatched at the same time to a place at the distance of 90 miles; the former by riding one mile per hour more than the latter arrived at the end of his journey one hour before him: find at what rate per hour each travelled. 20. A person rents a certain number of acres of pasture land for £70; he keeps 8 acres in his own possession, and sublets the remainder at 5 shillings per acre more than he gave, and thus he covers his rent and has £2 over: find the number of acres. 21. From two places at a distance of 320 miles, two persons A and B set out in order to meet each other. A travelled 8 miles a day more than B; and the number of days in which they met was equal to, half the number of miles B went in a day. Find how far each travelled before they met. 22. A person drew a quantity of wine from a full vessel which held 81 gallons, and then filled up the vessel with water. He then drew from the mixture as much as he before drew of pure wine; and it was found that 64 gallons of pure wine remained. Find how much he drew each time. 23. A certain company of soldiers can be formed into a solid square; a battalion consisting of seven such equal companies can be formed into a hollow square, the men being four deep. The hollow square formed by the battalion is sixteen times as large as the solid square formed by one company. Find the number of men in the company. 24. There are three equal vessels A, B, and C; the first contains water, the second brandy, and the third brandy and water. If the contents of B and C be put together, it is found that the mixture is nine times as strong as if the contents of A and C had been treated in like manner. Find the proportion of brandy to water in the vessel C. XXIX. Simultaneous Equations involving Quadratics. 264. We shall now solve some examples of simultaneous equations' involving quadratics. There are two cases of frequent occurrence for which rules can be given; in both these cases there are two unknown quantities and two equations. The unknown quantities will always be denoted by the letters x and y. 265. First Case. Suppose that one of the equation is of the first degree, and the other of the second depe Rule. From the equation of the first degree find the value of either of the unknown quantities in terms of the other, and substitute this value in the equation of the second degree. Example. Solve 3x+4y=18, 5x2-3xy=2. and then by substituting in the value of y we find that 266. Solve 3x2+5x-8y=36, 2x2-3x-4y=3. Here although neither of the given equations is of the first degree, yet we can immediately deduce from them an equation of the first degree. For multiply the first equation by 2, and the second by 3; thus 6x2+10x-16y=72, 6x2-9x-12y=9; therefore, by subtraction, 10x-16y+9x+12y=72-9; ; From this equation we obtain y= 19x-63 ; substitute this value in the first of the given equations; thus 3x2+5x-2(19x−63)=36; therefore therefore 3x2-33x+90=0; x2-11x+30=0. From this quadratic equation we shall find that x= 5 or 6; and then by substituting in the value of y we find that y=8 or 12. 267. Second Case. When the terms involving the unknown quantities in each equation constitute an expression which is homogeneous and of the second degree; see Art. 23. Rule. Assume y=vx, and substitute in both equations; then by division the value of v can be found. Example. Solve a2+xy+2y2=44, 2x2-xy+ y2=16. From this quadratic equation we shall obtain v=2 or 3. In the equation x2(1+v+2v2)=44 put 2 for v; thus x=2; and since y=vx, we have y=4. Again, in the same equation put 3 for v; thus x=√√2; and since y=rx, we have y = ±3/2. |