substituting the former value of y in the latter equation, 20x+15x=4200 therefore 35 x 4200 x=120 y=3x=360 11. There is a certain fraction, to the numerator of which 7 8 if 2 be added, the fraction becomes equal to; and if from 5 its denominator 2 be subtracted, it becomes equal to what 6 multiply the former equation by 8y, and the latter by 6(y-2), and they become, 8x+16=-7y From the first member of the former, transpose 16 to the other side; then multiply the former by 5, and the latter by 7; then they become 40x35 y-80 and by the third equation, 7y=8x+16=40+16=56 y=8 By numbering the equations [1], [2], [3], &c., and using the signs +, −, ×, ÷, the various operations may be more concisely expressed. The preceding steps of solution will then be represented thus:-- 12. If from the double of a certain number 6 be subtracted, the remainder is a number whose digits are those of the former in an inverted order, and the sum of the digits is 6, required the number. then 10 + y = the number (443) ; and therefore by the conditions of the question, 2 (10x+y)-6=10y+x by the former, 20 x+2y—6—10y+x 19 a = 8 y + 6 13. The sum of two numbers is s, and their difference is d: required the numbers. taking the difference 2y=s— d, and y = 12 (8 — d) This question was solved in another manner in the example preceding art. (249.) EXERCISES. 1. The sum of two numbers is 30, and their difference is 6: what are these numbers? Ans. 12, 18. 2. The sum of two numbers is 60, and their ratio is that of 2 to 3 required the numbers. Ans. 24, 36. 3. The difference between two numbers is 8, and twice the sum of their reciprocals is equal to three times their difference required the numbers. : Ans. 2, 10. 4. The sum of two numbers is 14, and 3 times the less is to 4 times the greater, as the square of the less to the square of the greater: what are the numbers? Ans. 6, 8. 5. Divide a line of 36 inches in length into two parts, whose ratio shall be that of 5 to 7. Ans. 15 and 21. 6. The ages of two persons are in the ratio of 3 to 4, but 10 years ago the ratio of their ages was that of 2 to 3: required their ages. Ans. 30 and 40. 7. Seven years ago, the age of a person A was just 3 times that of another B; and seven years hence A's age will just be double that of B's: what are their ages? Ans. 49, 21. 8. A person wished to distribute 3d. a-piece to some poor persons, but found he had not money enough in his pocket by 8d.; he therefore gave them each 2d., and found he had 3d. remaining required the number of poor people, and the money he had in his pocket? Ans. 11, and 25d. 9. A farmer wishing to purchase a number of sheep found that if they cost him 20s. a-head, he would be £2 short of money; but were they to be only 16s., he would then have £1 over: how many sheep were there, and how much money had he? Ans. 15, and £13. 10. A person A departs from a certain place, and travels at the rate of 7 miles in 5 hours, and 8 hours after another person B sets out from the same place, and travels at the rate of 5 miles in 3 hours: how long and how many miles does the first travel before he is overtaken by the second? Ans. 50 hours and 70 miles. 11. There is a certain number which is equal to 7 times the digit in the place of units, and if 18 be added to it, the sum is a number whose digits are those of the given number in an inverted order: what is the number? Ans. 35. 12. A cistern can be filled with water by two pipes running together in 6 hours, and the quantities discharged by the pipes in a given time are as 3 to 4: in what time would they separately fill the cistern? Ans. 14 and 10 hours. III. SIMPLE EQUATIONS CONTAINING THREE UNKNOWN QUANTITIES. (267.) The three methods given in the preceding case may be extended to the equations in this one, by properly modifying them; and a fourth method may also be sometimes advantageously used here. I. By Equating. (268.) 'Find a value of one of the unknown quantities in each of the three equations, and equate some of these values, so as to form two equations, by either placing the first equal to the second, and also to the third, or the second equal to the first, and also to the third; and as these equations will contain only two unknown quantities, their values may be found as in the former case.' The equations ought to be cleared of fractions, if necessary, before applying this rule. 1. Given EXAMPLE. 2x-3y+5%=15 3x+2y= 8 to find x, y, and z. -x+5y+2x=21) 15-2x+3y 5 z=3x+2y8 z= 2 Equating [4] and [5], 15-2x+3y =3x+2y-8 5 x-5y+21 and reducing this equation, 17x=55-7y... [7] Equating [5] and [6], 3 + 2 y - 8 = . 2 and reducing 5x=37-9 y ... [8] The equations resulting from the process of equating, namely, [7] and [8], are two equations containing two unknown quantities, a and g, whose values may be found by the rules of the former case. These values are thus found to be |