17 10 3 10 Ans. x=5, or ; and y=3, or Ex. 18. Given x2x2y+y2=49, to find the va and x* -2x3y3 +y'—x2 + y2=20, Ans. x=3, or±√6, or lues of x and y. (30±6 √5); and y=2, or-1, or 1(1±3/5).* Ex. 19. Given x-x3—3—y, and 4—x—y—y3, to find the values of x and y. Ans. x=4, or; and y=1, or 24. Ex. 20. Given x+x −4x2=y2+y+2, and xy= y2+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+2y3=60, to find the values of x and y. Ans. x=4/2, or F14; and y=+32. 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. Ans. x=2, or-; 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 x2y1-7xy =1710, and xy-y 12, to find the values of 30 and y. and y. -19 Ans. x=5, or, or 17±6-2 -15, or-6±√−2. and y=3, or Ex. 26. Given xy+xy=12, and x+xy=18, 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+y= =10, to find the values of x and y. Ans. x=3, or 1; or 4±√61; and y=1, or 3, or 41-61. Ex. 28. Given x2+4 √(x2+3y+5)=55—3y: and 6x-7y=16, to find the values of x and y. 51 x=5, or=31; Ans. x=5, or- ; or 7 46 9± √3895 -70+ √3895 49 Ly=2, or ; Ex. 29. Given x2+2x3y=441—x1y2, and xy= 3+x, to find the values of x and y. Ans. Sx-or-7; or−2± √ −17, y=2, or 4; or $4-17. Ex. 30. Given (x+y)2-3y=28+3x, 3x=35, to find the values of x and y. Ans. and 2xy+ (-255.) x=5, or, or-+V(-255) y=2, or 7. or Ex. 31. Given x2+3x+y=73—2xy, and y2+3y +x=44, to find the values of x and Ans y. Sx=4, or 16; or-12 √58, 4 44 Ex. 32. Given+136-2xy,and x+y=10, y2x2 to find the values of x and y. Ans. 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 √(21)+3. CHAPTER XI. ON THE SOLUTION OF PROBLEMS, PRODUCING QUADRATIC EQUATIONS. § 1. solution OF PROBLEMS PRODUCING QUADRATIC EQUATIONS, INVOLVING ONLY ONE UNKNOWN QUAN TITY. 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 guineas, 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. Let r the number; then 80 the price of each; x 80 80 -1, by the problem, and by reduction, x2+4x=320; ..x+4x+4=324, and x+2=±18; ..x=16, or -20. 80 And = 80 =5 guineas, the price of each, 30 16 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 their sum multiplied by the greater produces 266. What are those numhers? Let x= the greater; ..x-y= the less. and x.(2x-9)=266 ; .'. x2 9 266 <1111 2 2 9 47 4 4 completing the square, &c. x ...x=14, or—91; and x-9=5, or —18. 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 évery 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 x miles. x the number of miles B travelled per day; 19 - 19 361 19 extracting the root, -6=12; 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-x2=240-16x; .. x2-46x=-240. Completing the square, x2-46x+529=289; .*.x=23±17=40, or 6= lesser part; and 30-x=30-6=24= greater part. In this case, the solution of the equation gives 40 and 6 for the lesser part. Now as 40 cannot possibly be a part of 30, we take 6 for the lesser part, which gives 24 for the greater part; and the two numbers, 24 and 6, answer the conditions required. Prob. 5. Some bees had alighted upon a tree; at one flight the square root of half of them went away; at another eight-ninths of them; two bees then remained. How many then alighted on the tree? 16x2 Let 2x2=the number of bees; x+ +2=2x2, or 9x+16x2+18=18x2 ... 2x3—9x=18; (Art. 417), Multiplying by 8, 16x2-72x=144; adding 81 to both sides, 16x3-72x+1=225; ...4x=9±15=24, or -6; and x=6, or -1. ... 2x2=72, or4. But the negative value -1 of x, is excluded by the nature of the problem; therefore, 72= number of bees. 429. If, in a problem proposed to be solved, there are two quantities sought, whose sum, or difference, is equal to a given quantity, for instance, 2a; let half their difference, or half their sum, be denoted by x; then x+a will represent the greater, and xa the lesser, (Art. 102). According to this method of notation, the calculation will be greatly abridged, and the solution of the problem will often be rendered very simple. Prob. 6. The sum of two numbers is 6, and the sum of their 4th powers is 272. What are the numbers? 1 Let a half the difference of the two numbers; then 3+x the greater number, and 3-x= the lesser. .. by the problem, (3+x)*+(3—x)* =272, or 162+108x+2x=272; from which, by transposition and division, x4 +54x2=55; ... completing the square, x+54x2+729=784, and extracting the root, x2+27=±28; •*. x2=—27±28, and x=±1, or ±√-55. |