x + y = Regarding (x + y) as a single quantity, we have, by the rule, = − ↓ ± √1722 + } = − ¦ ± ¥. 41, and y = 42. Taking the first value of x+y, and combining with (3), we have, x= 25, y = 16. = = 12 14. x + y + √ x + y 23+ y3 = 189 From (1) taking only the first value, we have, 12+ 1 + √x + y ··· x + y = 9 Dividing (2) by (3), member by member, x2 xy + y2 = 21. Squaring (3), Subtracting (4) from (5), 15. x x2 + 2xy + y2 Combining (6) and (3), = 1722. y = 2 = 272 3xy = 60; or, xy = 20. x = 5, and y =: 4. (3.) (1.) (2.) (3.) (4.) (5.) (6.) (1.)} Raising (1) to 4th power, x-4x3y+6x2y2 — 4xу3 + y2 = 16 · Subtracting (3) from (2), and factoring, xy(4x2-6xy+4y2) = 256 . . (4.) Multiplying (1) by 2, squaring, multiplying by xy, and subtracting from (4), 2x2y2 = 256 x2y2+8xy = 128. xy = 4 ± √128 + 16 = xy = 8, xy or, or, Taking the first value xy 8, and combining with (1) x == 4, y = 2. y x 2x2y + y2 = 49 x2 +xy 7y2 + 1 = = = 11 30 66 y + 3x = 73 } 21. 23 + y3 } 16xy; + y2 = 20 } 4 ± 12; Ans. Ans. Ans. Ans. Ans. Ans. (3.) x = 4. y = 5, 23. y = 2. x = 3. = 2. x = 6. y = 5 x = 3. y = 1, x = 5, y = 4. or, = a2 } 3xy + 2x + y = 485 1 1 y x x2y = - xy2 = 16 PROBLEMS. (x + y) + xy x2 + y2 - (x + y) Multiplying equation (1) by 2, Adding (2) and (3), Ans. 2xy + 2(x+y)= 94 x2 + 2xy + y2 + (x + y) = 47 H = 62 = 156. (x + y)2 + (x + y) = 156 1. Find two such numbers, that their product, added to their sum, shall be 47, and their sum, taken from the sum of their squares, shall leave 62. Ans. Let x and y denote the numbers; then, from the conditions of the problem, Ans Ans. (a2 + 1). (a2 — 1). 10. • x= ĺ \ y = 15. Solving (4), and taking the first value of x + y, x + y = } + √156 + Substituting in (1), we have, x = have, Combining (5) and (6), Let x and y = 12 Cubing (1), we have, 5, and y = 7: the numbers required. x + y = 7 2. The sum of two numbers is 7, and the sum of their cubes is 91. What are the numbers? denote the numbers. Then = 343 we shall (5) x3 + 3x2y + 3xy2 + y3 (3.) Subtracting (2) from (3), factoring, and dividing by 3 xy(x + y) = 84 (4.) Dividing (4) by (1), member by member, xy = 12 Combining (1) and (5), we find, (6.) x = 3, and y = 4: the numbers required. (1.) ) (2.) (5.) 3. The fore wheel of a coach makes 6 revolutions more than the hind wheel, n going 120 yards; but if the circumference of each wheel be increased 1 yard, the fore wheel will only make 4 more revolutions than the hind wheel in going the same distance. What is the circumference of each wheel? Or, by reduction, Let x denote the number of yards in the circumference of the hind wheel, and y the number in that of the fore wheel. From the conditions of the problem, 9x 120 X 120 x + 1 20(x − y) Subtracting (1) from (2), + 6 = y2 +4 = = xy 120 y 120 Y+1 11y = 1; or, x = xy + 1 = y' = 4, and y": Substituting this value of x in (1), and reducing, 39 y = 11 20 11 x2 + y2 xy = x2 y2 3 Whence, Taking the positive value of y, and substituting in (1), we find, 11 x 5. Hence, the wheels are, respectively, 4 and 5 yards in circumference. 4. Find two numbers whose product is equal to the difference of their squares, and the sum of whose squares is equal to the difference of their cubes. Denoting the numbers by x and y, we have, (1.) { (2.) |