xy=6. + 2(2) to (1, 4elor y=2, 3, or −3Ħ√3. 14. x2+3x+y=73—2xy, ', & addAns. x=4, or 16; y2+3y+x=44. togeMr (1) *B! y=5, or -7. 17. x2+y2—x—y=78, ity x+y+xy =39. 2. Ans. x= a±√2b-a2 2 y= a=√2b-a2 Ans. x=2, or 16; y=2, or 1. Ans. x=9, or 3; y=3, or 9. 18. (x+y)2-3y=28+3x, 17 Ans. x=5, or 31; 19. x+y 3x xy-(x+y)=54. 20. x2+4(x2+3y+5)*—55—3y, Yom, Ans. x—5, or —53. ART. 251. 1. There are two numbers, whose sum multiplied by the less, is equal to 4 times the greater, but whose sum multiplied by the greater is equal to 9 times the less. What are the numbers? Ans. 3.6, and 2.4. 2. There is a number consisting of two digits, which being multiplied by the digit in tens place, the product is 46; but if the sum of the digits be multiplied by the same digit, the product is only 10. Required the number. Ans. 23. 3. What two numbers are those whose difference multiplied by the difference of their squares, will produce 32, and whose sum multiplied by the sum of their squares, is 272? Ans. 5 and 3. the sum of their Ans. 2 and 5. 4. The product of two numbers is 10, and cubes 133. Required the numbers. 5. If the sum of two numbers be multiplied by the greater, and that product be divided by the less, the quotient will be 24; but if their sum be multiplied by the less, and that product be divided by the greater, the quotient will be 6. Required the numbers. Ans. 4 and 8. -- NOTE. The preceding problems may be solved by pure equations. 6. The difference of two numbers is 15, and half their product is equal to the cube of the less number; find them. Ans. 18 and 3. 7. The product of two numbers is 24, and their sum multiplied by their difference is 20; find them. Ans. 4 and 6. 8. What two numbers are those whose sum multiplied by the greater is 120, and whose difference multiplied by the less is 16? Ans. 2 and 10. 9. What two numbers are those whose sum added to the sum of their squares is 42, and whose product is 15? Ans. 3 and 5. 10. Find two numbers such, that their product added to their sum shall be 47, and their sum taken from the sum of their squares shall leave 62. Ans. 5 and 7. 11. Find two numbers such, that their sum, their product, and the difference of their squares shall be all equal to each other. Ans. +5, and +5. 12. Find two numbers whose product is equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. Ans. 15, and 1(5+√5). " 13. A grocer sold 80 pounds of mace and 100 pounds of cloves for 65 dollars; but he sold 60 pounds more of cloves for 20 dollars than he did of mace for 10 dollars. Required the price of a pound of each. Ans. Mace 50 cts, cloves 25 cts. 14. A and B gained by trading 100 dollars. Half of A's stock was less than B's by 100 dollars, and A's gain was three twen tieth's of B's stock. Supposing the gains in proportion to the stock, required the stock and gain of each. Ans. A's stock $600, B's $400; A's gain $60, B's $40. 15. The product of two numbers added to their sum is 23; and 5 times their sum taken from the sum of their squares leaves 8. Required the numbers. Ans. 2 and 7. + 16. There are three numbers, the difference of whose differ- / ences is 5; their sum is 44, and continued product 1950; find the numbers. Ans. 25, 13, 6. 17. Divide the number 26 into three such parts that their squares shall have equal differences, and that the sum of those squares shall be 300. Ans. 14, 10, 2. 18. The number of men in both fronts of two columns of troops, A and B, where each consisted of as many ranks as it bad men in front, was 84; but when the columns changed ground, and A was drawn up with the front that B had, and B with the front that A had, then the number of ranks in both columns was 91. Required the number of men in each column. Ans. 2304, and 1296. ART. 252. FORMULE-GENERAL SOLUTIONS.- A general solution to a problem producing an equation of the second degree, like one of the first degree, gives rise to a formula (Art. 162), which expressed in ordinary language, furnishes a rule. We shall illustrate the subject by a few examples. 1. Two men, A and B, bought 300 (a) acres of land for 600 (b) dollars, of which A paid 300 (c) dollars, and B 300 (b−c) dollars. For certain reasons they agreed to divide the land so that B should pay 75 cents (d dollars) per acre more than A. How much land did each man get, and what did he pay per acre? GENERAL SOLUTION. Let x cost of A's land per acre, C b-c ··x+x+a, by the problem. Clearing of fractions and reducing, This formula gives the following rule for finding the amount paid per acre, by him who paid least per acre : RULE.- Find the cost of the whole number of acres at the difference between the prices per acre of the different pieces of land; subtract this from the amount paid for the whole land; square this remainder and add to it the cost of the whole number of acres at the difference between the prices per acre, multiplied by four times the sum of money paid by him who paid least per acre; extract the square root of this sum, add to the square root thus found, the remainder that was squared, and divide the sum by twice the whole number of acres; the quotient will be the amount paid per acre by him who paid least per acre. Having this, every other requirement in the question is easily found. For the particular case the results are, A paid $2.443 per acre, and got 122.8 acres nearly. 2. Investigate a formula for finding two numbers, x and y, of which the sum of the squares is s, and difference of the squares d. Ans. x=2(s+d); y=√2(s—d). 3. Investigate a formula for finding two numbers, x and y, of which the difference is d, and the product p. Ans. x=(d+√d2+4p); y={(—d+√d2+4p). 4. Investigate a formula for finding a number, x, of which the sum of the number and its square root is s. Ans. x=s+1−√s+1. 5. The same when the difference of the number x, and its square root is d. Ans. x=d+1+√dt. 6. Given x+y=s, and xy=p, to find the value of x2+-y2, x3+y3, and x1+y1, in terms of s and p. Ans. x2+y2-s2-2p; x+y=s3-3ps; x+y=s^-4ps2+2p2. Let the student express each of the preceding formulæ in the form of a RULE, and exemplify its use, by forming examples with particular numbers, and then solving them. NOTE. In the great variety of equations that occur, which may be solved as equations of the second degree, it is not to be supposed that Rules can be given for every operation necessary for their solution. The artifices by which algebraic calculations are abridged, are numerous, and their successful application can be learned only by practice. In the following article, which is intended only for advanced students, we shall exhibit some of these artifices. ART. 253. SPECIAL SOLUTIONS AND EXAMPLES.-If an equation can be placed under the form (x+a)X=0, in which X represents an expression involving x, the unknown quantity; since the equation will be satisfied by making either factor =0, we have x+a=0, and X=0. Therefore, x=-α, is one solution of the equation, and the other values of x will be found by solving the equation X=0. Hence, whenever an equation is simplified by division, or the omission of a factor, if the divisor or factor contains the unknown quantity, one solution, at least, of the equation will be found by putting that divisor or factor equal to 0. Thus, the equation x3-x2-4x+4=0, may be placed under the form (x-2)(x2+x-2)=0. Hence, x-2=0, or x=+2, and from the other factor we find x=+1, or −2. The difficulty to be overcome in applying this artifice, consists in finding the factors of the given equation, or in transforming it so that it can be readily separated into factors. 2 Also, √a—1=3, by dividing by √ä+1. Whence, √x=2, or -1; and x=4, or 1. 2. x3-3x=2. (Add 2x to each side.) Ans. x=—1, or 2. |