Ratio of Sum of two first Terms to that of two last. Moreover, in finding these sums and differences, each antecedent may be multiplied by any number, provided its consequent is multiplied by the same number. 72. Corollary. These rules may also be applied to the proportion A: C = B : D obtained from means, and give and A: B = C: D by transposing its A+B:C+D=A−B: C-D = mA+nB: mC+nD=mA-n B: mC-nD A: CB: D; A + B : A − B = C + D : C — D mA+nB: mA-n B=mC+nD: mC-nD; that is, the sum of the first two terms of a proportion is to the sum of the last two, as the difference of the first two terms is to the difference of the last two, or as the first term is to the third, or as the second is to the fourth. Likewise, the sum of the first two terms is to their difference, as the sum of the last two is to their dif ference. Moreover, in finding these sums and differences, both the antecedents may be multiplied by the same number, and both the consequents may be multiplied by any number. 73. Two proportions, as and A: BC: D E: FG: H, Ratio of Reciprocals. may evidently be multiplied together, term by term, and the result A X EB X F = C × G: DX H is a new proportion. 74. Likewise, all the terms of a proportion may be raised to the same power. 75. Theorem. The reciprocals of two quantities are in the inverse ratio of the quantities themselves. 1 such that the product of the first A and the last is the A 1 same with that of the second B and the third each pro B duct being equal to unity. Letters used for unknown Quantities. CHAPTER III. Equations of the First Degree. SECTION 1. Putting Problems into Equations. 76. The first step in the algebraic solution of a problem is the expressing of its conditions in algebraic language; this is called putting the problem into equations. 77. No rule can be given for putting questions into equations, which is universally applicable. The following rule can, however, be used in most cases, and problems, in which it will not succeed, must be considered as exercises for the ingenuity. Represent the required quantities by letters of the alphabet. Perform or indicate upon these letters the same operations which it is necessary to perform upon their values, when obtained, in order to verify them. It is usual to represent the unknown quantities by the last letters of the alphabet, as v, w, x, y, z. EXAMPLES. The following problems are to be put into equations. 1. A person had a certain sum of money before him. From this he first took away, the third part, and put in its Examples of putting Questions into Equations. stead $50; a short time after, from the sum thus increased 4 he took away the fourth part, and put again in its stead $70. He then counted his money, and found $120. What was the original sum? Method of putting into equations. Let x= the original sum expressed in dollars. After taking away the third part and putting in its stead $50, there remains two thirds of the original sum increased by $50, or x + 50. If from this sum is taken a fourth part, there remains three fourths; to which is to be added $70, giving (x+50)+70 =x+107; which is found to be equal to $120. We have, therefore, for the required equation, 2. A merchant adds yearly to his capital one third of it, but takes from it at the end of each year $1000 for his expenses. At the end of the third year, after deducting the last $1000, he finds himself in possession of twice the sum he had at first. How much did he possess originally? the original capital in dollars, the required Ans. If x equation is 3. A courier, who goes 31 miles every 5 hours, is sent from a certain place; when he was gone 8 hours, another was sent after him at the rate of 22 miles every 3 hours. How soon will the second overtake the first? Solution. If x = the required number of hours, the number of hours which the first courier is on the road is x+8; Examples of putting Questions into Equations. and the distance which he goes is obtained from the proportion 5: x+8= 31 distance gone by 1st courier; whence, by art. 64, distance gone by 1st courier = £8 (x+8). The distance gone by the second courier is obtained from the proportion whence 3:x= 221: distance gone by 2d courier; distance gone by 2d courier = 15 x. But as both couriers go the same distance, the required equation is f8(x+8)= Y x. 4. A courier went from this place, n days ago, at the rate of a miles a day. Another has just started, in pursuit of him, at the rate of b miles a day. In how many days will the second courier overtake the first? Ans. If x = the required number of days, the required equation is b x = a (x + n). 5. A regiment marches from the place A, on the road to B, at the rate of 7 leagues every 2 days; 8 days after, another regiment marches from B, on the road to A, at the rate of 31 leagues every 6 days. If the distance between A and B is 80 leagues, in how many days after the departure of the first regiment will the two regiments meet? the required number of days, the required Ans. If x equation is Zx + 3(x-8)=80. |
