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Examples of Substitution of Unknown Quantities.

5. Solve the two equations

or

x3 y - x2 y2 + xy2 = 156,

ху (x3 - у3) - 2 x2 y2 (x - y) + (x - y)2 = 157. Ans. x = 4, or = - 3, or = ( 1±√−51);

x=

y = 3, or = + 4, or = + (−1±√−51);

7*))+781,

x=±√(-1572±2√(624+157*),
y=±±√(−1572±2√(624+157*))*-78.

6. What two numbers are they, whose difference is 1,

and the difference of whose third powers is 7?

Ans. 1 and 2, or - 2 and 1.

7. What two numbers are they, whose difference is 3, and the sum of whose fourth powers is 257?

Ans. 4 and 1, or 4 and - 1, or((-79)+3) and (±√(-79)-3).

159. When the first member of one of the equations, reduced as in art. 88, is homogeneous in regard to two unknown quantities, the solution is often simplified by substituting for the two unknown quantities, two other unknown quantities, one of which is their quotient.

The same method of simplification can also be employed when such a homogeneous equation is readily obtained from the given equations.

EXAMPLES.

1. Solve the two equations

x2-6xy + 8 y2 = 0,

x2y +6xy2+8y3+(x-2y) (y2-5y+4)=0.

Examples of Substitution of Unknown Quantities.

Solution. Retaining the unknown quantity y, introduce

instead of x, the unknown quantity q, such that

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from which the given equations become

q2 y2 - 6qy2 + 8 y2 = 0,

q2y3-6qy3+8y3+(qy-2y) (y2-5y+4)=0.

Both these equations are satisfied by the value of y,

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But if we divide the first of these equations by y2, and the

second by y, we have

q2 - 6q + 8 = 0,

2

q2y2-6qy2 +8y2 + (q-2) (y2-5y+4) = 0;

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being substituted in the other equation, reduces the first member to zero, and therefore y is indeterminate; that is, x and y may have any values whatever, with the limitation that x is the double of y.

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Examples of Substitution of Unknown Quantities.

Solution. 13 times the first equation, diminished by the second equation, is

12 z5 + 13 x3 y2 - 4 x y = 0;

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12 q5 y5 + 13 q3 y5 - 4 qy5 = 0.

Which is satisfied by the value of y,

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and this value of y, being substituted in the given equations, produces

x5

5,

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whence

12 q* + 13 q2 — 4 = 0,

q = ± 1, or q = ±√-.

Now the first of the given equations becomes, by the substitution of

x = q Y,

95 y5 + q3 y5 = 5;

hence, by the substitution of the above values of q, we have

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5

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or y=±√ × √-3,x=√12+ × √2 = √20.

Examples of Substitution of Unknown Quantities.

3. Solve the two equations

81 x4 + 9 x2 y2 = 20 y4,

(y2-y)2+(3xy+2y)2-9x2(2y+3)-12y(x+2y)=0.

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5. What two numbers are they, twice the sum of whose squares is 5 times their product, and the sum of whose sixth powers is 65. Ans. 2 and 1, or - 2 and - 1.

6. What two numbers are they, the difference of whose fourth powers is 65, and the square of the sum of whose squares is 169. Ans. 2, and ± 3.

Form of any Equation.

CHAPTER VI.

General Theory of Equations.

SECTION I.

Composition of Equations.

160. Any equation of the nth degree, with one unknown quantity, when reduced as in art. 80, may be represented by the form

1

Ax + Bxn-1 + Cxn-2 + &c. + M = 0. If this equation is divided by A, and the coefficients represented by a, b, &c., m, it is reduced to

BC
M
-, &c.,
A' A'
A

xn + axn-1 + bxn-2 + &c. + m = 0.

161. Theorem. If any root of the equation

xn+axn-1 + bx-2 + &c. + m = 0

is denoted by x', the first member of this equation is divisible by x- x'.

or

Demonstration. Denote x - x' by x[1], that is,

x[1] = xx',

x = 1 + 2[1].

If this value of x is substituted in the given equation, if Pz[1] is used to denote all the terms multiplied by x[1], or

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