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found by the same process. The other three equations may be solved in the same

manner.

4. These equations may be solved by the same method of elimination at these in 3.

5. Since

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y and x"=y", x=y' and x=;
=y", .'. y=y, and
and x-ny

Substitute this value of x in the second equation,

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6. Divide each side of the equation by x, then x2-9=2·

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which can be solved by a quadratic equation, from which x

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-1 and -2.

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a2-b2

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a2 + b2

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number. If a be odd, a will be fractional, if a be even, x will be integral.

a =

Suppose ma — 2a = (m − a)2, then a2+4

4

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(3) If a=1, 2x2 -1 is a square when x=1, 5, &c.

(4) Here a + bx + cx2 is a square, if 4a x cx2=(bx)2. Also when b2-4ac is a square, a+b+ca2 may be resolved into two factors.

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(5) Is a particular example of this case, for a=6, b-13, c=6, b2 - 4ac = 25.

Here the factors are 2+3x and 3+2x, and x=

In order that x may be positive

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There are other cases which may be considered when a is a square number, also when c is a square number, and the cases in which the expression cannot be made a square.

7y-24, and any assumed value of ม will

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2. (1) Since 5x+7y-2y=xy, .'. x= give a corresponding value of x. If the equation be restricted to positive values of a and y. Let y=1, 2, 3, 4, 5, 6, 7, &c.

It will be found that the corresponding values of x are

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When y−1, x=1; y=5, 2-, these values of x and y must be rejected, as

not being positive integers.

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(7) The equation 2x(x+y) + y2 – b(y − 3) = 0 may be reduced to the form y2+2(x-3)y=-18-2x2.

(8) x=1, 3, 4, 7; y=39, 19, 15, 9.

(10) The least values are x=70, y=30.

(9) x=4, 5, 6; y=27, 11, 5.

3. Let x, y denote the numbers, then x2 + y2 is to be a square number. Assume 2xy x2 + y2 = (nx − y)2=n2x2 − 2nxy + y2, then x= ; and for all values of x, x2-1

corresponding values of y may be found. If integral values of x and y only be required. If y=x2 -1, then x=2, and by taking x=2, 3, &c., a series of corre sponding values of y will be found.

Next let x2 - y2 = (x−ny)2, and by a similar process, if y = 2n, then x=a2 +1. 8. Let the numbers be x2, y2, z2, then x2+2=2y. Let x=m+n, and y=m-n, then x2+z2 = 2(m2 +n2)=2y2, and m2+n2=y2, the question is reduced to find m and n, the sum of whose squares shall be a square. See example 4. The numbers 1, 25, 49, and 4, 100, 196, are two integral solutions of the problem.

9. Let x denote the number of men in the side of the first of the five hollow squares, then x+5, x+10, x+15, +20 are respectively the numbers of men in a side of the other four hollow squares.

Then x2(x-10)2=number of men in the first hollow square.

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The whole number of men is 100x + 500. If y denote the number in the side of the solid square, then y2 = 100x+500, and it is required to determine x, so that 100(x+5) may be a complete square.

ELEMENTARY ARITHMETIC, WITH BRIEF NOTICES OF ITS HISTORY.

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EACH section of the Arithmetic may be purchased separately; also the twelve sections together, done up in boards, with cloth covers, at 48. 6d.

ELEMENTARY ALGEBRA,

WITH BRIEF NOTICES OF ITS HISTORY.

SECTION IX.

RATIO, PROPORTION, AND VARIATION.

BY ROBERT POTTS, M.A.,

TRINITY COLLEGE, CAMBRIDGE,

HON. LL.D. WILLIAM AND MARY COLLEGE, VA., U.S.

LONDON:
LONGMANS AND CO.,

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