3. A cistern can be filled by two pipes in 12 hours, and by the first of them alone in 20 hours: in what time would it be filled by the second alone? Put x for the time required. 12 in 12 hours; and would be the quantity supplied by 20 the first in 12 hours. But in twelve hours the two running together filled the cistern, 12 12 .. = 1, X 20 and multiplied by 20x, 240+12x=20x, Rule 3d; by transposition, &c. 240 8x, Rule 1st; 4. How much rye, at four shillings and sixpence a bushel, must be mixed with 50 bushels of wheat, at six shillings a bushel, that the mixture may be worth five shillings a bushel? Let x be the number of bushels of rye. Then 9x the price of the rye in sixpences. 600 the price of the wheat in sixpences. (50+)10 the price of the mixture. 9x+600-500+10x by the question. Hence 100 by transposition. 70. SCHOLIUM. The transferring of problems into algebraic equations will be facilitated by studying carefully the following remarks:-1st, If the sum of two numbers be given, and one of the numbers be represented by x, then the other will be the sum minus x. 2d, If the difference of two numbers be given, and the less be represented by x, the other will be x plus the given difference; and if the greater be represented by x, the other will be x minus the difference. 3d, If the product of two numbers be given, and one of them be represented by x, the other will be the product divided by x. 4th, If the quotient of two numbers be given, and one of them be represented by x, the other will be the quotient multiplied by x. 5th, If the sum of two numbers be represented by s, and the less number by x, then the difference of the numbers will be s—2x; and if the greater be represented by x, their difference will be (2x-s). 19. What number is that which being increased by its half and its third part, the sum will be 154? Ans. 84. D 20. What number is that to which, if its third and fourth parts be added, the sum will exceed its sixth part by 17? Ans. 12. 21. At a certain election, 311 persons voted, and the successful candidate had a majority of 79: how many voted for each ? Ans. 195 and 116. 22. What number is that from which, if 50 be subtracted, the remainder will be equal to its half, together with its fourth and sixth parts? Ans. 600. 23. A husband, on the day of his marriage, was thrice as old as his wife, but after they had lived together 15 years, he was only twice as old: what were their ages on the marriage day? Ans. Husband's, 45; wife's, 15. 24. It is required to divide L.300 between A, B, and C, so that A may have twice as much as B, and C as much as A and B together: what was the share of each? Ans. A's, L.100; B's, L.50; and C's, L.150. 25. A cistern can be filled with water by one pipe in 12 hours, and by another in 8: in what time would it be filled by both running together? Ans. 4 hours. 26. Two pieces of cloth, which together measured 40 yards, were of equal value, and the one sold at 3s., and the other at 7s. a-yard: how many yards were of each? Ans. 12 yards at 7s., and 28 at 3s. 27. A has three times as much money as B, and if B give him L.50, A will have four times as much as B: find the money of each. Ans. A's, L.750; B's, L.250. 28. After 34 gallons had been drawn from one of two equal casks, and 80 from the other, there remained thrice as much in the first as in the second: what did each contain when full? Ans. 103 gallons. 29. A person paid a bill of L.100 with half-guineas and crowns, using in all 202 pieces: how many pieces were there of each sort? Ans. 180 half-guineas, 22 crowns. 30. There is a cistern which can be supplied with water from three different pipes; from the first it can be filled in 8 hours, from the second in 16 hours, and from the third in 10 hours: in what time will it be filled if the three pipes be all set running at the same time? Ans. 3 hours 2819 minutes. 31. Solve the above question generally on the supposition that the first pipe can fill the cistern in a hours, the second in b, and the third in c. Ans. abc ab+ac+bc SIMULTANEOUS EQUATIONS. 71. When two or more unknown quantities are to be determined, there must always be as many independent equations as there are unknown quantities; and since the values of these unknown quantities must be the same in all the equations, the values are said to subsist simultaneously, and the equations are called simultaneous equations. CASE I. To determine two unknown quantities from two independent equations. 72. RULE I. Make the coefficients of one of the unknown quantities the same in both equations, then by adding or subtracting these equations, there will result an equation containing only the other unknown, whose value may be found by the previous rules. NOTE 1. The equations are to be subtracted when the quantity whose coefficient is rendered the same in both equations, has the same sign in each, and added when it has opposite signs. NOTE 2. The coefficients of either of the unknown quantities may always be rendered the same in both equations, by multiplying the first equation by the coefficient of the unknown quantity, which is to be made to disappear in the second equation, and the second equation by the coefficient of the same letter in the first. By this means the coefficients of that quantity will evidently be the same in both, for it will be the product of its two coefficients in the original equations. 73. RULE II. Find a value of one of the unknown quantities in terms of the other from each of the equations, and make these values equal to each other, which will give an equation containing only the other unknown, from which its value can be found by the previous rules. 74. RULE III. Find a value of one of the unknown quantities in terms of the other from one of the equations, and substitute this value instead of it in the other, from which there will again result an equation containing only one unknown quantity, which may be solved as before. 75. RULE IV. Multiply one of the equations by a conditional quantity n, then add this product and the other equation together: let n fulfil the condition of making the coefficient of one of the unknown quantities 0, then the equation will only contain the other unknown; determine the value of n that fulfils the above condition, and substitute this value instead of it in the resulting equation, and the value of the other will be determined. NOTE 3. All the above operations can be performed first in relation to one of the unknown quantities, and then in relation to the other, which will give independent values of each; or the value of one of the unknown quantities being found, its value can be substituted instead of it in either of the given equations, from which the value of the other can be determined. EXAMPLE. Given 3x+5y=35, and 7x-4y-19; to find the values of x and y. From (1.) by transposition and division, x= 35-5y 3 245-35y=57+12y by mult. by 21. And substituting this value of y in (1.) we obtain 3x+20=35, Hence 3x=15, And x=5, as before. By Rule 3d. It has already been found from (1), that x= stituting this value, instead of x in (2) it becomes 7(35—5y)—4y=19. 245-35y-12y=57, by mult. by 3. In the same manner may x be found from either equation, by substituting a value of y found from the other equation. By Rule 4th. Multiplying (1) by n, and adding (2), we obtain (3n+7)x+(5n—4)y=35n+19, which, if the coefficient of y be 0, becomes (3n+7)x=35n+19; and since the -oefficient of y is =0, 5n-4=0; hence n=3; substitutng this value of n in the equation (3n+7)x=35n+ 19, it 9x-47. ecomes Hence 47x=235. 7 Next, let n be such as to render the coefficient of x=0, Then the equation will become (5n—4)y=35n+19; and ince 3n+7=0.. n=-3, substituting this value instead of n in the equation, and changing the signs, it becomes 1. Given 153y=623. Hence 47y=188. And .. y=4. (3x+2y=56 to find the values of x and y. 12x+3y=545 Ans. x 12, y=10. 2. Given 5x-7y=81 to find the values of 3. Given 4. Given XC and y. Ans. x=10, y=6. x+y=7 to find the values of x and y. 4x-y=34) x+y=95 3x+y=38 to find the sx+2y=12 J I x2—y2=201 to find the Ans. x=9, y=8. values of x and y. =12, y=4. Ans. x= values of x and y. Ans. x=6, y=4. to find the values of x and y. Ans. x=1(s+d), y={(s—d). 5. Given 6. Given x+y=s | x-y=d 3x+2y +2x-16 2x-3 2x+7y + -4 9 11 to find the values of x and y. Ans. x=6, y=3. to find the values of XC and y. Ans. x= cb'-c'b ca'-c'a ́ab'—a'b''—ba'—ab" The above equations may all be solved, by substituting in the answer to the (8) the proper values of a, b, c, a', b', and c', with their proper signs; only (7) would require to be reduced to the proper form before the substitution can be made. CASE II. 76. To determine the values of three or more unknown quantities from as many independent equations as there are unknown quantities. |