Now, let x+y=5, and x5+y=1056; then by substituting 3 for a, and 1056 for b, in the formulæ of roots, the values of x and y will be found; that is, x=3±1, or 3± √-19; and y=31, or 3 √ — 19. Or, by substituting the above values of a and b in the equation 10az-20a z24-2ab, it becomes 30+540z+486+1056; from which the values of z may be found; whence, by substitution, the values of x and y will be determined, as before. Ex. 11. Given x+4y-14, and y3+4x=2y+11, to find the values of x and y. Ans. x=-46, or 2; and y=15, or 3. Ex. 12. Given 2x+3y=113, and 5xa—7y3 =4333, to find the values of x and Ans. x=35, or y. 3899 3268 and y=16, or17 17 Ex. 13. Given x+4y2-256-4xy, and 3y2-x2 =39, to find the values of x and y. Ans. x=±6, or ±102; and y=±5, or ±59. Ex. 14. Given x"+y"=2a”, and xy=c2, to find the values of x and y. Ex. 15. Given x2+2xy+y2+2x=120-2y, and xy-y2=8, to find the values of x and y. Ans. x 6, or 9, or-95; and y=4, or 1, or-3±√5. Ex. 16. Given x2+y2-x-y=78, and xy+x+y =39, to find the values of x and y. Ans. x-9, or 3; or-6-39; and y=3, or 9, or-61√−39. Ans. x=5, or ; and y=3, or- 10 Ex. 18. Given x1—2x3y+y3=49, to find the vaand x-2xy + y2x2 + y2=20, lues of x and y. Ans. x+3, or±√6, or ±√(30±6 √5); and y=2, or-1, or 1(1±3.√√5).* I 2 Ex. 19. Given x-x2: =3-y, and 4-x find the values of x and y. x=y—y3, to Ans. x=4, or ; and y=1, or 21. I Ex. 20. Given x+x-4x=y2+y+2, and xy= y+3y, to find the values of x and y. Ans. x=4, or 1; and y=1, or—2. Ex. 21. Given x2+xy=56, and xy+2y=60, to find the values of x and y. Ans. x4/2, or 14; and y=3/2, or ±10. Ex. 22. Given x-y=15, and xy=2y3, to find the values of x and y. Ans. x=18, or 12; and y=3, or-21. Ex. 23. Given 10x+y=3xy, and 9y-9x=18, to find the values of x and y. 3 Ans. x=2, or-1; and y=4, or §. Ex. 24. Given x+y: x-y :: 13:5, to find the and y2+x=25, S values of x Ans. x=9, or-14; and y=4, or-61. Ex. 25. Given a2y-7xy 1710, and xy-y 12, to find the values of x and y. and y. Ans. x=5, or, or -15, or-6±√-2. -19 = 17±6-2 and y=3, or 3 Ex. 26. Given xy+xy2=12, and x+xy318, to find the values of x and y. Ans. x2, or 16; and y=2, or . *There are four other values, both of x and y, which are all imaginary, Ex. 27. Given x+y+√(x+y)=6, and x2+y2 =10, to find the values of x and y. Ans. x=3, or 1 ; or 4±√—61; and y=1, or 3, or 411-61. Ex. 28. Given x2+41 (x2+3y+5)=55—3y, and 6x-7y=16, to find the values of x and y. Ex. 29. Given x2+2x3y=441-x'y', and xy= 3+x, to find the values of x and y. Ans. (x=8, or-7; or-2±√-17, y=2, or 4; or $17. Ex. 30. Given (x+y)2-3y=28+3x, and 2xy + 3x=35, to find the values of x and y. Ans. x=5, or 1, or-±√(−255) y=2, or 3, or—11= √(−255.) Ex. 31. Given x2+3x+y=73-2xy, and y2+3y +x=44, to find the values of x and y. x=4, or 16; or—127 158, y=5, or ·7; or-1± 158. Ans Ex. 32. Given to find the values of x and Ans. 136-2xy,and x+y=10, y. Sx=6, or 4; or 5±5√(−1), y=4, or 6; or 575√—(}}). Ex. 33. Given y1-432-12xy2, and y2=12+2xy, to find the values of x and y. Ans. x 2, or 3; and y=6, or 1(21)+3. CHAPTER XI. ON THE SOLUTION OF PROBLEMS, PRODUCING QUADRATIC EQUATIONS. § 1. SOLUTION OF PROBLEMS PRODUCING QUADRATIC EQUATIONS, INVOLVING ONLY ONE UNKNOWN QUANTITY. 428. It may be observed, that, in the solution of problems which involve quadratic equations, we sometimes deduce, from the algebraical process, answers which do not correspond with the conditions. The reason seems to be, that the algebraical expression is more general than the common language, and the equation which is a proper representation of the conditions, will express other conditions, and answer other suppositions. Prob. 1. A person bought a certain number of oxen for 80 gui. neas, and if he had bought four more for the same sum, they would have cost a guinea a piece less; required the number of oxen and price of each. and by reduction, x2+4x=320; ..x2+4x+4=324, and x+2=±18; 80 80 And x 16 5 guineas, the price of each. The negative value (-20) of x, will not answer the condition of the problem. Prob. 2. There are two numbers whose difference is 9, and theif sum multiplied by the greater produces 266. What are those num bers? Let x= the greater; x-y= the less. and x.(2x-9)=266 ; .'. x2. 9 266 x= 9 47 x- =- ; 4 4 completing the square, &c. x .'. x=14, or —91; and x-9=5, or —181. Here both values answer the conditions of the problem. Prob. 3. A set out from C towards D, and travelled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went every day one-nineteenth of the whole journey; and after he had travelled as many days as he went miles in one day, he met A. Required the distance of the places C and D. Suppose the distance was a miles. 19 and also = the number of miles B travelled per day; he met A. the number of days he travelled before 7x x2 361 +32+ 19 19=8, =8, or 4; and x=152, or 76, both which values answer the conditions of the problem. The distance therefore of C from D was 152, or 76 miles. Prob. 4. To divide the number 30 into two such parts, that their product may be equal to eight times their difference. Let x= the lesser part; ... 30-x= the greater part, and 30-x-x, or 30—2x= their difference. Hence, by the problem, x(30-x)=8(30-2x), or 30x-x=240—16x; .'. x2-46x=-240. Completing the square, x2-46x+529=289; ..x=23±17=40, or 6= lesser part |