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The value of x will be easily found to be ±2, or

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==±8 1/1/1/1

14. Find the values of x and y in the Equations 3y2x2-39, and x2+4xy=256—4y2.

Substituting vy for x in these Equations

=

=±1; or y = 31

39
256
3-v2 v2+4v+4

Clearing this Equation of fractions,

39v2+156v+156-768-256v2.

6

5'

From this Equation we shall find v= We can now easily find y=±5, or 59. 6

Then, x=5x =±6; or x=59×

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3y2-v2y2-39, and v2y2+4vy2-256-4y2.

Finding the values of y2 from these Equations, and equating them, we have

or

Y
Clearing the second Equation of fractions,

x3+y3=18xy.

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102

59

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Solutions by Means of Two Auxiliary Unknown Quantities.

15. Find the values of x and y in the Equations

x2 y2
x+y=12, and + =18.

x

Assuming xv+z, and y=v-z, we shall have (x+y)=(v+2)+(v−z)=2v=12.

Hence v 6.

Then x=6+z, and y=6-2.

Substituting these values of x and y in the third Equation, we have

(6+z)3+(62)3—18(6+2) (6—z).

Developing both sides of this Equation, and reducing,

54z2 216; hence z=±2. Then since x=6+z, we have x=6±2=8, or 4. And since y=6-2, we have y=6=2=4, or 8.

16. Find the values of x and У in the Equations x+y=10, and x3+y3=280.

Assuming xv+z, and y=v-z, we have x+y=2v 10. Hence v=5.

Then x= 5+%, and y=5-2.

Substituting these values of x and y in the second Equation, (5+2)+(5-2)3=280.

Developing the first member of this Equation, and reducing,
302-30; hence z=±1.

Since x=6+z, we have x=5±1=6, or 4.
Since y=6-2, we have y=5+1=4, or 6.

17. Find the values of x and y in the Equations x+y=11, and x2+y1=2657.

Substituting v+z for x, and v-z for y in the first Equation, and uniting similar terms, we find

2v 11, or v=

11

2

11

11

Therefore x=

+z, and

2

y= -2.
2

Substituting these values of x and y in the second Equation,

11

11

4

(+)+(-)=2657.

1

Developing each term of the first member, of this Equation, by the method of the Binomial Theorem, we have

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Adding these two Equations together, and recurring to the preceding Equation, we find

29282 1452%2

16

4

+

Clearing this Equation of fractions,

2+6

have

22224-2657.

This Equation, by developing the two binomials in the first member, becomes

2=

Then being equal to

And y being equal to

(1)

2

72

29282+580822+3224-42512.

From this Equation, omitting imaginary roots, we shall have

19

3

= √//i3

4

2

32

2

11

2

11

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+224-2657.

=+

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18. Find the values of x and y in the Equations,
x+y=10, and 25+y5-17050.

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Substituting v+z for x, and v-z for y in the first Equation, and adding similar terms,

2v=10; or v=5.

Therefore x=5+z, and y=5—z.

Substituting these values of x and y in the second Equation,

(5+2)+(5-2)5-17050.

Developing each binomial in the first member of this Equation, we

(5+z)5=3125+3125z+1250z2+250z3+25z4+25.

(5-2)5-3125-3125z+125022-25023+2524-25.

Adding these Equations together,

(5+2)+(5-2)56250+2500z2+5024.

Hence we shall have

6250+2500z2+50z4—17050.

From this Equation, omitting imaginary values, we shall find z=±2. Hence x=5±2=7, or 3; and y=5=2=3, or 7.

Miscellaneous Solutions and Exercises.

19. Find the values of x and y in the Equations xy=6, and x2+x=18-y2-y.

From the second Equation, we have
x2+y2+x+y=18.

Adding twice the first Equation,

x2+2xy+y2+x+y=30.

This Equation may be expressed by

(x+y)2+(x+y)=30;

which is quadratic with reference to x+y, and from which we shall find x+y=5; or x+y=-6; hence

x=5—y; or x=—6—y.

Substituting these values of x, successively, in the first Equation, we have

5y-y2 6, and —6y—y2—6.

The first of these Equations will give

y=3, or 2; and the second will give y=-3±√√3.
6-y; we have x=5-3=2, x=5—2.

Since x-5-y; or x= ==3, or x=-6—(−3±√/3)=-3√/3.

20. Find the values of x and Y in the Equations x+y=6, and x2y2+4xy=96.

The second Equation may be expressed by

(xy)+4(xy)=96.

This Equation is quadratic with reference to xy, and will give

xy=8, or -12; hence

8

12

Y

y

Substituting the first value of x in the first Equation,

8

+y=6; whence y=4, or 2.

y

Substituting the second value in the same Equation,

12

+y=6; whence y=3±√/21.

y

By substituting each of these values of y for y in the first Equation we shall find the corresponding values of x to be 2, 4, and 3√21.

x or

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21. Find the values of x and y in the Equations

x*y*—2y2, and 8x*—y3=14.

Dividing the first Equation by y2, we find

8x

x3-2y; whence y* —

Substituting this value of y* in the second Equation,

=14.

2

Clearing this Equation of its fraction,

16x-2=28.

From this Equation (263), we shall find

. .

= 2

14, or 2; hence x=2744, or 8.

Substituting each of the values of x3 in the second Equation,
a
112-y=14, and 16—y2— =14.

From the first of these Equations,

=98; hence y=9604.

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