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-A'C for the first remainder; and if we write the terms of the equations in a contrary order, multiply the second by C and then divide the product by the first, we shall get (AC' A'C)x+BC'B'C for the first remainder.

Hence, eliminating a from these remainders, we shall get the same equation of condition as before.

3. Because we may in like manner eliminate an unknown letter and its powers from any two algebraic equations, it is clear that we shall have another method of elimination, which is called Elimination by the method of the greatest Common Divisor. On account of the striking analogy between this method and that of Addition and Subtraction, we do not think it important to take any further notice of it.

IV. ELIMINATION BY ARBITRARY MULTIPLIERS.

RULE.

Multiply all the equations but one by separate arbitrary multipliers, and add the resulting equations and the remaining equation together. Then (in the sum) put the coefficients of all the unknown letters but one (in the given equations) separately equal to 0, and the given equations will be reduced to a single equation which contains one of the unknown letters (of the given equations) and the multipliers; consequently, after the multipliers have been found from the assumed equations, the letter may be found from the reduced equation.

EXAMPLES.

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1. Given, 5x8y = 4, 11x-10y= 14, to find x and y. Multiplying the terms of the first equation by m, and adding the corresponding terms of the second, we get (5m + 11)x — (8m + 10)y: 4m + 14.

=

Assuming 8m + 100 or m = —

5 4'

we shall have x =

- 4m + 14

5m + 11

4; and if we assume 5m +11=0 or m=

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2. Given, 3x+2y= 12 and 4x + 3y = 17, to find x and y. Multiplying the terms of the first by m, and adding the

corresponding terms of the second, we shall have (3m + 4)x

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3

= 3; and assuming 2m + 3 = 0 or m = — we have x=

2'

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Remark.-If we assume 12m + 170 or m =

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2m + 3

we

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shall get (3m + 4)x + (2m + 3)y = 0 or

2

3

= the ratio of x to y; consequently, if we know y we can easily get x, and vice versa.

3. Given, 4x-3y+22=14, 6x + 7y - 112 = 9x+2y+32= 62, to find x, y, and z.

25, and Multiplying the terms of the first by m, those of the second by n, and adding their corresponding terms and those of the third, we get (4m + 6n + 9)x + (− 3m + 7n + 2)y + (2m - 11n+3)= 14m + 25n+62.

Assuming 4m+6n+9=0,−3m+7n+2=0, we get m=

35

46; consequently, z=

14m+25n+62
2m-11n+3

= = 3.

Assuming 4m+6n+90 and 2m - 11n + 3 = 0,

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From the values of y and 2, we shall, from either of the given equations, get a = 5; or we can get a by assuming the 3m + 7n + 2 = 0, 2m — 11n + 30, which 14m + 25n+ 62

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Since this method of elimination is not often used in practice, the preceding examples must suffice for its illustration.

V. PARTICULAR METHODS OF ELIMINATION, ILLUSTRATED BY

EXAMPLES.

1st.-1. Given, 3x+2y=68 and 2x + 3y = 72, to find x and y.

Multiplying the terms of the first by 72, and those of the second by 68, we shall get (3x+2y) × 72 = 68 × 72 and (2x + 3y) × 68 = 72 × 68; consequently, since the second members of these equations are equal, we may equate their first members, and shall thence get (3x+2y) × 72 = (2x + 3y) × 68 or (3x + 2y) × 18 = (2x + 3y) × 17; which is

easily reduced to 54x + 36y=34x + 51y or x =

3y

4

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Putting for a in either of the given equations, we get

3y 4

y=16, and thence x =

3y 4

gives x = 12.

Remarks.-1. If we put xyz, and substitute

yz for x in the given equations, and eliminate y from the resulting equations, we shall get an equation in 2, whose solution will

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sequently, our method comes to the same thing as to put yz for x in the given equations, and then to eliminate y and find z from the resulting equation.

2. It is easy to perceive that we may, in a similar way, solve any number of simple equations.

2. Given, x + y = 7 or x2 + 2xy + y2 = 49 and x2 + y2 : 25, to show how to find x and y.

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Here we have (x2 + 2xy + y2) × 25 = (x2 + y2) × 49, or

25ay 12x2+ 12y; which is easily reduced to

x

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2 25/x

12 y

=1, and putting =2, we have the equation 2 22

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25

12

=-1, whose solution will give 2, and thence x and y can easily be found.

3. "Given, x2 + xy=56 and xy + 2y2 = 60, to find x and y."

Here we have (x2 + xy) × 60 = (wy + 2y3) × 56 or 15a2 +

15xy = 14xy + 28y2; which is easily reduced to

=

28 15'

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2

+

()

1/x) 15 y 2 28

()

=2 we have the equation 22 +

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Solving the equation we get 2, and thence we easily obtain the values of y and x.

It is hence clear if we have two equations which contain two unknown letters, as x and y, such that each of them is homogeneous in terms of x and y, that by putting yz for x in the equations, and eliminating y from them, we shall get an equation whose solution will give 2, and thence y and x can be found.

2d.-1. Given, x+y=7 and x2 + y2 = 25, to find a and y.

Assume 2+v and y = 2-v, then, by substitution, the equations will be reduced to z + v + 2 − v = 7 and (z+ v)2 + (≈ — v)2 = 25, or to z = and 2v2 +222 = 25.

7 2

7

25

Putting for z in the second equation, we get v2 =

2

2

49 1 4 = 4'

1

or taking the square root, we have v=

Hence,

2

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Taking the half sum and half difference of the preceding

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49

1

equations, we get 22 =

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4

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Remark.-Because x z + v and y = 2-v give z =

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2

2

Y , and that xyz-v2, we shall get

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wy = (1 + ?)" — (~ ~ Y)" ; consequently, we perceive how to

y 2

reduce any product to the difference of two squares. Thus, since 209 = 19 × 11, if we put 19 for a and 11 for y, we get x + y

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4; consequently, we shall have 209

=15-42; and it is clear that any odd integer (except 1) may be regarded as being the difference of the squares of two integers.

3. Given, (x − y) (x2 — y3) = a and (x + y) (x2 + y2) = b, to find x and y.

Putting 2+v and y = 2-v, we get x - y = 2v, x2- y2 = 4zv, x+y=2z, and a2 + y2 = 222 + 2v2; consequently, substituting these values in the given equations, they will be reduced to 8zv2= a and 423 +4zv2 = b. Subtracting the first of these equations from twice the second, we get 823 = 2b-a, whose cube root gives 22 26 — a; consequently, dividing the first by four times this value of

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a

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It is clear from the preceding examples that this method of elimination will generally be very useful in solving two equations which contain two unknown letters, as x and y, such that they are symmetrically involved in the equations; or (which is the same) when the equations remain the same, when a is put for y, and y for a, in them.

3d.-1. Given x+y=7 and xy= 12, to find x and y. Squaring the first and subtracting four times the second from it, we get x2 + 2xy + y2 — 4xy = x2 - 2xy + y2 = (xy) = 1, whose square root gives x - y = 1.

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Hence, we easily get x = 4 and y = 3, since the half sum and half difference of the equations + y = 7 and x-y= 1 will clearly give the values of x and y.

2. Given, xy=7 and xy = 60, to find x and y.

Here we have (x − y)2 + 4xy = (x + y)2 = 289, or x + y =

17; consequently, x = 12 and y = 5.

3. Given, x+y=a and a3 + yb, to find x and y.

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· xy + y2

=

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a

x + y

have (x + y)2 = x2 + 2xy + y2 = a2; consequently, by sub

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