Again, multiply the first equation by 5, then will but hence, by subtraction, Eq. (1), multiplied by 7, and subtracting, we get From (1) we get and thence 2. Let x+2y 5x+5y+10 z 66 z = = 95, = = = = x = 19 36 - 31 = 11, and 3x 5, 6. 9, to find the values of x, y, and z. From the second equation, 2y + 4z but (1) and subtracting, we get Subtracting this last equation from the third equation, gives Ans. x = 12, y = 18, z = 9. 3. 5x+3y=84, 4x+6z=108, 5y+7z110. 5. 3x+4y+5x=38, 4x+3y-4 z=1, 6x-2y+7= = 34. 6. x+y+z = 15, x + y − z=3, x − y + z = 5. Ans. x 4, y = 5, z= 6. - 7. 3x+2y-z 20, 2 x + 3y+6z=70, x − y +6 z = 41. Ans. x=5, y = 6, z= 7. х x =8+ y z 212 + = 10. 6' 2 3 Ans. x 12, y = = 12, z = 12. 24 = C. 2 a b c 4 7 5 3 5 + - + y y 2 8 X Ans. x 6, y = 12, z = = 8. = 2= a b + b c - a c' 2 a b c abac - bc. QUESTIONS PRODUCING SIMPLE EQUATIONS. 99. When a question is proposed to be solved by the principles of algebra, its meaning ought to be perfectly understood, and its condition or conditions exhibited in the clearest manner. If there is only one unknown quantity to be determined, one condition will be sufficient, and if several quantities are to be determined, we must have as many independent conditions as there are unknown quantities to be found. In reducing problems to algebraical equations no general rule can be given, but the following hint may be useful. Suppose that the values of the quantities to be determined are actually found, and consider by what operations the truth of the solution may be verified. Then, instead of known numbers, make use of unknown symbols for the quantities to be determined, and let the same operations be performed with these letters as were performed with the numbers, and we shall obtain certain equations expressing the conditions of the question. The solution of these equations will give the values of the unknown quantities. Thus if the question were to find the number which exceeds its fourth part by 27, then supposing the number were found, viz. 36, the process of verification would be as follows: Now if x denote the number required, we should, by following the same steps, arrive at the equation and the solution of this equation would give the number required; for, multiplying by 4, we get 4 x x = 108, or 3 x = 108; hence x = 36. 100. When several quantities are to be determined, there must be given an equal number of independent conditions, and for the purpose of expressing these conditions we may find it convenient to satisfy one or more of them by a judicious selection of an unknown quantity which has some known relation to the required quantity. Thus, if the sum of two quantities be given equal to 2s, we may denote their difference by 2x; then s+x and sx would denote the greater and less numbers. In a similar manner, if the difference of two numbers be given equal to 2d, then x+d, and x d might be employed to denote the two numbers, for the condition of their having a given difference would be fulfilled. Also, if the ratio of two numbers be given, as m to n, then mx and nx might be put for the numbers, and one condition would be fulfilled; and if the product of two numbers be given equal to p, then P Р x and might represent the two numbers, for then x x = p = the given product, and so on. 1. What number is that to which if 8 be added, one-third of the sum is equal to 5. Let x represent the number required, then if 8 be added, the sum is x+8, and one-third of this sum is of the question, we have x+8 = x+8 Hence, by the condition 5; therefore x + 8 = 15, and x = 7. 2. What two numbers are those whose sum is 48, and difference 18. Let x denote the less number, then x + 18 will denote the greater, and one condition is fulfilled; and we have only to satisfy the other by adding the two expressions together; thus 30; = = x + x + 18 48, or 2 x 48 18 = hence x = greater number. Or thus, by two unknown Quantities. Let the greater number and y = the less; then the two conditions of the question expressed in algebraical language are 3. A bill of 25 guineas was paid with crowns and half-guineas; and twice the number of half-guineas exceeded three times that of the crowns by 17; how many were there of each? Let x denote the number of crowns, and y the number of half-guineas, then the second condition of the question gives And as we have employed two unknown quantities, we must have another condition to furnish a second equation. The other condition is obviously this: crownsy half-guineas = 25 guineas. And here it is necessary to reduce these different coins to the same denomination, and thus divest the terms of the equality of their concrete character. The highest common denomination is sixpences; the value of a crowns is therefore 10 x sixpences, the value of y half-guineas is 21 y sixpences, and 25 guineas is 25 × 42 1050 sixpences; hence the preceding equality becomes in the abstract = 10 x + 21 y 1050 ..... (2). = To resolve these equations, multiply (2) by 3, and (1) by 10, then we have hence, by adding, we get 83 y 30 x = 3320, and y = 40, the number of 1780 17 21, the number of crowns. These numbers will half-guineas. Also from (1) we get 3 x 63; therefore x be found to fulfil both the conditions expressed in the question. 4. A garrison of 1000 men was victualled for 30 days; after 10 days it was reinforced, and then the provisions were exhausted in 5 days: find the number of men in the reinforcement. Let x denote the number of men in the reinforcement; then since the consumption varies directly as the product of the men and time, we have ог 1000 × 30 = 1000 x 10 + (1000 + x) × 5 5. There is a number consisting of two digits, which, when divided by their sum, gives the quotient 4; but if the digits be reversed, and the number thus formed be increased by 12, and then divided by their sum, the quotient is 8; find the number. = Let x = the digit in the ten's place, and y = the digit in the unit's place; then 10 x + y the number, and 10 y + x = the number when the digits are reversed; hence the two conditions of the question furnish these two equations: 10x + y x + y 10 y + x + 12 Clearing these equations of fractions, we get 10 y + x + 12 but = 2 y 4, y = == hence x = 2 x = 8, and the number is 10 x + y = 6. Two couriers, A and B, start at the same time from two towns, C and D, distant c miles from each other, and travel in the direction of the straight line passing through C and D. Now supposing A travels at the rate of a miles, and B at the rate of b miles per hour, when and where will they be together? = Let A and B travel in the same direction from C and D towards M, the place where the couriers will be together. Let CM x, and D M = y; then we have x y = C. Also since A travels x miles at the rate of but these times are equal by the condition of the question; hence, To apply this result to particular examples, let the distance C D = 10 miles = c, and let a of x and y we get = 8, and 67; then from the preceding values = which are the spaces travelled by A and B before they come together. If a = 16 and b = 15, then x = 160, and y 150 miles; hence it appears that the nearer their rates of travelling approach to equality the greater will be the spaces to be travelled before they come together. And if the rates of travelling are equal, then x = ∞ and y = ∞ ; whence it appears that they cannot come together, as is manifest from the equality of their rates of travelling. = 130. Now if we suppose b greater than a, then a b would be negative, and the values of x and y would be negative. Thus if a = 12 and b = 13, then x = 120 and y To interpret the meaning of these negative results, we may observe that, if the couriers travel in the direction CDM, they can never come together, for B travels at a greater rate than A does, and therefore the distance between them is continually increasing. The meaning of the negative result will be easily understood, by considering the couriers to have been travelling at the specified rates towards C and D from some point M' where they had been together. Travelling from M' in the direction M'CD at the rates of 12 and 13 miles respectively, A would have travelled 120 miles and B 130 miles before their distance apart had been 10 miles; hence if M' be 120 miles from C and 130 from D, the couriers, starting together from the point M', and travelling at the proposed rates, would be simultaneously at C and D. The true interpretation of the negative results 120 and 130 is, that they denote distances in a direction directly opposite to those contemplated in the question. 8 and y 2, then will x = - 2; and 2 signifies that B is travelling in a direction opposite to that of A, so the result y 2 corresponds to such a supposition, and denotes that the place of meeting M" is 2 miles to the left of D and 8 miles from C towards D. = 8, and b Lastly if a as the supposition b = = = = EXAMPLES FOR PRACTICE. 1. What number is that which exceeds its sixth part as much as 26 exceeds its fourth part? Ans. 24. 2. What number is that to which its third part being added, the sum shall be equal to its half added to 10? Ans. 12. 3. Divide 1000 into two parts, so that one of them shall be threefifths of the other? Ans. 375 and 625. 4. If 2 be added to the numerator of a certain fraction, its value will be; and if 2 be added to the denominator, its value will be; what is the fraction? Ans. 1 5. A farmer rents 150 acres of lands for 2377. 10s., part at 25s. and part at 45s. per acre; how many acres are there of each kind? Ans. 100 acres at 25s., and 50 at 45s. 6. A person after spending 107. more than a fifth of his income, had remaining 351. more than half of it; what is his income? Ans. 1501. 7. In a naval engagement, one-third of the fleet was taken, one-sixth sunk, and two ships were burnt. In a storm after the action, oneseventh of the remainder was lost, and only 24 ships are left; of how many ships did the fleet consist? Ans. 60 ships. |
