Key to Davies' Bourdon, With Many Additional Examples, Illustrating the Algebraic Analysis; Also, a Solution of All the Difficult Examples in Davies' Legendre

Cover Key to Davies' Bourdon, With Many Additional Examples, Illustrating the Algebraic Analysis; Also, a Solution of All the Difficult Examples in Davies' Legendre
Key to Davies' Bourdon, With Many Additional Examples, Illustrating the Algebraic Analysis; Also, a Solution of All the Difficult Examples in Davies' Legendre
Charles Davies
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a;2 - y^ + ar - y = 150 • • ' • (1), x^ + y^ + x + y = SSO • • • (2); adding member to member, and reducing, x^ + x = 240; whence, x=-^± ^^40 + 1 = - - =b y ; or, considering only the positive solution, a; = 15; whence, from (1), by substitution, y = 9.
8. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number : what are the numbers ?
Let X and y denote the numbers.
From the conditions of the problem, a; — y = 15 • • • ♦ (1), |^ = y3, or ^
... = 2y2 . . . (2); substituting in (1) and dividing both members by 2, whence, ^ 2"" 2 ' 1 /15 , 1 1 11 2^ = 4^V-2+l6 = 4=^T^ considering only the positive solution, y = 3 ; whence, from (1), ar = 18.
182.] EQUATIONS OF THE SECOND DEGREE. 91 9. What two numbers are those whose sum multiplied by the greater, is equal to 77 ; and whose difference, multiplied by the lesser, is equal to 12?
Let X and y denote the numbers.
From the conditions, ' (x + y) xz= 77, or x^ + xy-=n . • • (1) ; (a:-y)y=12, or ary - y2 = 12 • • • (2); make x z=py; whence, {p^+p)y^ = 17, or, y*=^5^ • • • (3), 12 {p -l)y» = 12, or, y^=j^ry • • • (4); equating the second members and reducing, 2 65 77 ^ -I2^=-12' 65 / 77 , 4225 65 23 whence, ^ = - ± y'- _ + -^ = _ ± _; taking the upper sign, _88__ 11 -^ "" 24 "■ 3 9 /0£t Q substituting in (4), y = V "q" = 2 '^' whence, ^ = "o" V^j 42 21 taking the lower sign, -^ — 24 ~ l2 ' 92 KEY TO DAVIE8* BOURDON.


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