-- 47. (7x-21)-(4-1(31-5x)) = 181 = 5(1 x — 10). 48. 4.709 — 1(5.7 x — 31) — 0.3(2‡ — 5.3x) = 0. 49. 5-21(4.6-31x) = 4.7x-0.8(3x-1). 50. 51. 5.7x-21(7.8—9.3 x) = 5.38—43(0.28 +3 6 x). 52. 5.05x-505(505 — 5.05 x) = 50.5 x — 50.5(50.5 x — 5.05). +8_35r+16) = x- + 5 – 28 5x =7+ Show that the equation is impossible. 6 + 2 3 5 5x 2x Show that this equation is satisfied for any values that may be assigned to x. CHAPTER II PROBLEMS WHICH LEAD TO SIMPLE EQUATIONS OF ONE 178. In the present chapter the methods already given will be applied to the solution of problems, in order that the student may understand their practical application. In a problem certain quantities are given, and others, which have certain assigned relations to the first quantities, are to be determined. The method of solving the problem may be thus described in general terms: Represent the unknown quantities by letters, and express in algebraic symbols the relations which hold between the unknown quantities and the given quantities; then equations will be obtained from which the values of the unknown quantities may be determined. In the present chapter only problems which may be solved by using one unknown quantity will be discussed. 179. PROBLEM 1.-Divide 53 quarters between two persons so that the first shall have one-third more than the second, plus four quarters. Let x = the number of quarters belonging to the second person. Then x+x+4= the number of quarters belonging to the first person. The sum of the quarters belonging to the first and second persons will be equal to 53 quarters. and, removing the denominator 3 by multiplying both members of the equation by 3, or, ... finally Hence 6x+x=147, 7x=147 x21, number of quarters belonging to second person. + 4 = 21+7+4 =32, number of quarters belonging to first person. 180. PROBLEM 2.-How much money is there in a purse the sum of whose fifth and fourth parts is $225? Let x the number of dollars in the purse. fifth of dollars in the purse, and Then = one one-fourth of dollars in the purse. But a fifth and fourth part of the number of dollars in the purse is $225; hence, Remove the denominators by multiplying the equation by 20 and get 181. PROBLEM 3.-Two persons have the same capital; the first lends his at five per cent, the second lends his at three per cent. The revenue of the first exceeds that of the second by $400. Find the capital. Recall the principle established in Arithmetic: To find the interest on a given capital, multiply the capital by the rate and divide the result by 100. Let x be the capital desired. The revenue of the first person will be that of the second Since the revenue of the 5x 100 3x 100 first person exceeds that of the second by $400, the equation follows, To solve this equation transpose the terms involving x to the first member, Then the capital sought is $20000. Verification. The interest on the capital, $20000, at 5 per cent is $1000; and at 3 per cent is $600. The first income of $1000 exceeds the second income, $600, by $400, 182. PROBLEM 4.-The denominator of a fraction exceeds its numerator by 2; and, if 1 is added to both numerator and denominator, the resulting fraction will be equal to . What is the fraction? Let x = the numerator of the fraction. Then x2 the denominator of the fraction. If 1 is added to both the numerator and the denominator, the Verification. The numerator 3+1=4; the denominator 5+1=6, and the new fraction will be 23 as required. 3+1 4 = 183. PROBLEM 5.-If A can do a piece of work in 8 days and B in 10 days, in what time will they perform it together? x Let the number of days required. If A can perform the work in 8 days he can perform of the work in one day, and B can perform of the work in one day; and consequently A and B working together can perform + of the work in one day. But if they both can perform the work together in x days, they can together perform of the work in one day. Thus, 184. PROBLEM 6.-Find the number of passengers who were in a train leaving New York, under these conditions: the train lost of the passengers at the first station, 12 passengers at the second station, of the remaining passengers at the third station, and of the remaining passengers at the fourth; 42 passengers continued their journey from the fourth station. |