and third b, and the sum of the squares of the second and third c. Prob. 16. The sum of two numbers is 16, and the sum of their cubes 1072. What are those numbers? Ans. 7 and 9. Prob. 17. The sum of two numbers is 2a, and the sum of their cubes is 26. To find the numbers. Prob. 18. Two magnets, whose powers are as 4 to 9, are placed at a distance of 20 inches from each other. It is required to find, on the line which joins their centres, the point where a needle would be equally attracted by both; admitting that the intensity of magnetic attraction varies inversely as the square of the dis It is required to point which is Prob. 19. Two magnets, whose powers are as m to n, are placed at a distance of a feet from each other. find, on the line which joins their centres, the equally attracted by both. Prob. 20. A set out from C towards D and traveled 6 miles an hour. After he had gone 45 miles, B set out from D towards C, and went every hour of the entire distance; and after he had traveled as many hours as he went miles in one hour, he met A. Required the distance of the places C and D. Ans. Either 100 miles or 180 miles. Prob. 21. A set out from C towards D and traveled a miles per hour. After he had gone b miles, B set out from D towards C, and went every hour 4th of the entire distance; and after he had traveled as many hours as he went miles in one hour, he met A. Required the distance of the places C and D. (186.) On the solution of quadratic equations containing two unknown quantities. An equation containing two unknown quantities is said to be of the second degree, when it involves terms in which the sum of the exponents of the unknown quantities is equal to 2, but never exceeds 2. Thus are equations of the second degree. The general solution of two equations of the second degree containing two unknown quantities, depends upon the solution of an equation of the fourth degree containing one unknown quantity. Hence the principles already established are not sufficient to enable us to solve all equations of this description. Yet there are many particular cases in which they may be reduced either to pure or affected quadratics, and the roots determined in the ordinary manner. Multiply the second equation by 2, we have, 2xy=56. (187.) It will sometimes facilitate the operation to substitute for one of the unknown quantities, the product of the other by a third unknown quantity; which method may be applied to advantage whenever the sum of the dimensions of the unknown quantities is the same in every term of the equation. See Article 30. Ex. 2. Given (x2 + xy = 56 { xy + 2y2 = 60} to find the values of x and y. Here, agreeably to the above observation, let x=vy, then v2y2+vy2=56 vý2 + 2y2 60. Whence from the first of these equations y' = 56 v2 + v Therefore by equating the right hand members of these two expressions, we shall have The values of x are read minus or plus 14, while those of y are read plus or minus 10; showing that when the value of y is +10, x must be taken 14 in order to satisfy the original equations; and when y is taken 10, x will be equal to +14. (188.) A solution may sometimes be shortened by substituting for the unknown quantities, the sum and difference of two other quantities; which method may be used when the unknown. quantities in each equation are similarly involved. Ex. 3. Given x2 y2 y to find the values of x and y. x+y=12 Here according to the above observation assume x=z+v, = and y――v. Then by adding these two equations together, we shall have x+y=2z = 12 or z=6. Also since x=6+v, y=6—v, and by the first equation x3 + y3 = 18xy, we shall obtain by substitution (6 + v)3 + (6 — v)3 = 18 (6 + v) (6 — v), or by involving the two parts of the first member, and multiplying those of the second From the second equation x = 2y3. Substituting this value in the first equation, we have 2y2y= 15. y. (189.) In many cases it may be convenient to solve the equation first, considering one of the quantities as known; when the |