Note. This problem, we see, admits of a double solution; for whether 12 and 19 or-5 and 2 be taken as the corresponding less and greater parts, the conditions of the question are answered by each pair of numbers. Illus. Thus the difference of 12 and 19...... That of 5 and 2.. Half product of 12 and 19 + 30 = is 7. is 7. 5 and 2 +30= + 30 = 25 = 2 =square of the less. -572 =square of the less. EXAM. IV. To divide the number 30 into two such parts that their product may be equal to eight times their difference. And 30-x=30-40 or 30 - 6-10 or 24 = greater part. Here the equation gives for the less part 40 and 6; but as 40 cannot possibly be a part of 30, we take 6 for the less part which allows 24 for the greater; and the two numbers 24 and 6 answer the conditions of the question. EXAM. V. A merchant bought cloth for 331. 15s. which he sold again at 21. 8s. per piece, and gained by the bargain as much as one piece cost him: what did he gain by the bargain? Solu. Let the number of pieces. 675 Then = the number of shillings which each piece cost; and 48x= the number of shillings which he sold the whole for. Therefore 48r - 675 what he gained by the bargain. Bat it is obvious the number required is 15, because the conditions of the question are such that there cannot be a negative or fractional number of pieces. EXAM. V. A and B set off at the same time to a place at the distance of 150 miles from that which they left; A travels 3 miles an hour faster than B, and arrives at his journey's end 8 hours 20 minutes before him: it is required to find what rate each person travelled per hour. Also X 150 x+3 the rate per hour at which B travels. the rate per hour at which A travels. the number of hours for which B travels. the number of hours for which A travels. But A arrives 8 hours 20 minutes (8 hours) sooner at his journey's end than B ; Hence +8= 150 150 25 150 3 150 And completing the square x2 + 3x + 2 = 54 + 225 = 9 miles an hour for B. - 6 miles an hour for A. Since this is a question of motion, it is evident that if motion to a place be reckoned positive, motion from the same place must be reckoned negative : But A and B are moving to a given place, therefore the positive numbers 6 and 9 are taken for the respective rates per hour of A and B. EXAM. VI. It is required to divide the number 14 into two such parts that their product shall be 50. - Solu. Let x = one part; then 14 x= the other part; And ro-14x + 49 —— 50+ 49= −1, or x − 7 = + √−1; But both the values of x are impossible, because the question itself is impracticable. OF QUADRATIC EQUATIONS, WITH TWO UNKNOWN QUANTITIES. 76.) For the solution of Quadratic Equations with two unknown quantities, the three following methods, which are generally used, will be found applicable to a vast variety of questions. The FIRST METHOD is performed by this general (77.) Rule.-Find the value of one unknown quantity in terms of the other, in one of the given equations, and substitute this value for it in the other equation; from which there will arise a quadratic equation soluble by the methods already given. Solu.-Multiply by 3, and from the first equation 2x+y=27. Hence, 3ry=3(27—3)y=210; Or, 3x (27-y) xy=420, multiplying by 2. That is, 81y-3y' =420. Or,.. 27y-y=140, dividing by 3 ..y-27y=-140, changing the signs. 2. A certain number consists of two digits; the left hand digit is equal to three times the right hand digit; and if we subtract 12 from the number itself, the remainder will be equal to the square of the left hand digit; it is required to find the number. Solu.-Put x=left hand digit 7 then will 100+y be the number. Hence, by the question, r=3y, and 10ry 12=x® ; Therefore, 30y+y-12=9y2 by substitution; And 3y = 3 x 39; consequently, the number is 93. 3. Let there be two numbers such that, if the less be taken from 3 times the greater, the remainder shall be 35; and if 4 times the greater be divided by 3 times the less plus 1, the quotient will be equal to the less number; to find these two numbers. Ans. 13 and 4. it is re 4. There is a certain number, and the sum of its digits is 15; now if 31 be added to their product, the digits will be inverted ;. quired to find that number. Ans. to find the values of x and y. 78. (78.) The SECOND METHOD: When those parts of the given equations that contain the two unknown quantities bear a particular relation to one another; as, for instance, their sum and difference; their sum and product; their sum and the sum of their squares; their product and the sum of their squares; their sum and the sum of their cubes, &c. &c. The equations may then be managed by some one of the following solutions. Then, the square of the first equation, x2+2ry+y2=a® (A) And the second equation x by 4 By subtracting (B) from (A) = = 4xy =46 (B). Extracting now the square root but .. a+√a-4b Hence, by addition, 2x=a+√√a2-4b, or x 2 And, by subtraction, 2y=a√a2-4b, or y= (81.) 3. Let x+y=a x2 + y2=b Then the square of the first equation is x+2xy + y2=a2 (A); From which take Therefore But multiply by 2, and then we have 4xy=2a2—2b (B). Subtract (B) from (A), and x2-2xy +y3—a2—2a2—2b=2b—a®. Hence, -y=±√2b—a2. Having now r+y and x-y, we should proceed by the first Example (Art. 79) to find the values of x and y. Therefore x+2xy + y2=a+2b, or x+y=+√a+26. Hence, x and y may now be found as in the preceding exa (83.) 5. Suppose that x+y=a and x+y=b Then will x+3x2y+3xy2+y3=a, cubing the first equa Subtract +y=b, the second equation |