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223. These are the simplest cases that arise, but they are specially important since the solution in a large number of other cases is dependent upon them.

As a rule our object is to solve the proposed equations symmetrically, by finding the values of x+y and x-ý. From the foregoing examples it will be seen that we can always do this as soon as we have obtained the product of the unknowns, and either their sum or their difference. Example 1. Solve x2 + ya=74

(1), xy=35

(2). Multiply (2) by 2; then by addition and subtraction we have

x2 + 2xy + y2=144,

28 – 2xy +ya= 4; Whence

2+y= +12,

2 Y= £ 2. We have now four cases to consider; namely, x+y=12, x+y= 12, x+y=-12, *+y=-12,1 X - Y= 2.S

2 y=-
2.

2.

2-y=- 2.) From which the values of x are 7, 5, -5, -7; and the corresponding values of y are 5, 7, -7, -5. Example 2. Solve 22 + y2=185

.(1), x+y=17

(2). By subtracting (1) from the square of (2) we have 2xy=104;

.(3). Equations (2) and (3) can now be solved by the method of Art. 222, Example 1; and the solution is

=13, or 4,1

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.: wy= 52.

EXAMPLES XXVIII. a. Solve the following equations : 1. +y=28, 2. x+y=51, 3. x+y=74, XY=187. xy=518.

xy=1113. 4. X-y=5, 5. X-y=8, 6.

xy=1075, XY=513.

X - Y=18. 7. xy=923,

8. X-Y= -8, 9. x-y= -22, x+y=84.

Xy=1353.

Xy=3848.

XY=126.

11. X - = -18,

10.

XY = - 2193,

x+y= -8. 13. x2 + y2=89,

xy=1363.

12.

XY = - 1914, x+y= -65. 15. x2 + y2=65,

14. x2 + y2=170,

XY =40.

XY =13.

XY=28.

27.

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xy=12,

16.2.2 + y2 =178, 17. x+y=15, 18. X - Y=4,

x+y=16. x2 + y2=125. 22+y2=106. 19.2 + y2=180, 20. 202 + y2=185, 21. 20+y=13, X - Y=6.

X Y=3.

32 + y2 =97. 22. x+y=9, 23. X Y=3, 24. 22 - xy + y2 = 76, x2 + 2y+ya=61. 2:2 – 3xy + y2 = - 19.

x+y=14. 1 1 1

1 1 7 25. 1o (x-3)=1, 26.

+-=2,

+ y

y

12'
22 – 4xy + y2 =52. x+y=2.
28. ax+by=2, 29. x2 + pxy+y=p+2,
abxy=1

9.29 + xy +9y9=29+1.
224. Any pair of equations of the form
x2+pxy + y2=a2.....

.(1), x+y=6

.(2), where

P
is

any numerical quantity, can be reduced to one of the cases already considered; for by squaring (2) and combining with (1), an equation to find xy is obtained; the solution can then be completed by the aid of equation (2). Example 1. Solve 23 Y3=999

2C - Y=
3

(2). By division, x2 + xy + y2=333

x2 – 2xy + y2 = 9; by subtraction, 3xy=324,

(4). From (2) and (4)

x=12, or

9,7

y= 9, or Example 2. Solvexc4 + ^y2 + y4=2613..

.(1), 2c2 + xy + y2= 67..

(2). Dividing (1) by (2) 22 xy + y2= 39..

(3). From (2) and (3) by addition, 2 + y2 =53; by subtraction,

xy=14; whence

x= +7, +2,1

[Art. 223, Ex. 1.] y=+2,

(3);

from (2)

xy=108

12.S

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3 y=3, or

2" EXAMPLES XXVIII. b. 1. 2-3+3=407, 2.2.3+y=637, 3. x+y=23, X +y=11.

X+y=13. 23+73=3473. 4. 23 =218, 5.

X-Y=4, 6. 203 y3 =2197, X-Y=2. 23 43=988.

X y=13. 7. 24+24y2 + y4=2128, 8. 204 + x-ya+y4=2923, 22 + xy+y2=76.

22 2y+y2=37. 9. x4+x2y2 +y4=9211, 10. 24+2ya+y=7371, .22 - xy + y2=61.

22 - xy + y2=63. 1 1 481

1 61

y 11. 12.

13. + -21, 32576'

y2 900'

y
1 1 29

x+y=6.
+
Y

24
Y

34

15
+
15.

16. .2.3-y=56, y

xo+y2 XY x-y=4.

22+xy+y2=28.

x+y=8. 17. 4 (22+y^)=1724,

18.

23+y3=126, x-y=6.

22 - xy + y2=21.

+

.22

+

22

XY=30.

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HOMOGENEOUS EQUATIONS OF THE SAME DEGREE.

225. The following method of solution may always be used when the equations are of the same degree and homogeneous. Example. Solve 22 + xy + 2y>=74.

.(1), 2x2 + 2xy + y2=73.

(2). Put y=mx, and substitute in both equations. Thus 22 (1 + m + 2mo)=74.

(3), and x(2 + 2m + m2)=73.

.(4).
1+m+2m2 74
By division,

2+2m + m2 73
.:: 73+73m +146m2=148+148m + 74m2;
... 72m2 75m 75=0,

24m2 - 25m - 25=0);
.. (8m+5) (3m – 5)=0;

5 5

or

.. m =

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**(1-%+69)

=74;

–64;

5 (i) Take m= and substitute in either (3) or (4).

5 50 From (3)

8' 64

64 x 74 .. x2=

74 .. x= +8;

5 .: y=mx=

X= 75.

8 5

5 50
ii) g+

=74,
74 x9
22=
74

5 .. X=+3; .: y=mx=3&=5. The student will notice that, having found the values of X, we obtained those of y from the equation y=mx, using, in each case, the value of m employed in obtaining those particular values of x.

=9;

5+4y.

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i

226. When one of the equations is of the first degree and the other of a higher degree, we may from the simple equation find the value of one of the unknowns in terms of the other, and substitute in the second equation. Example. Solve 3x – 4y=5

.(1), 32c2 — xy - 3y2=21..

(2). From (1) we have

3

3(5+4y)2 y(5+4y) and substituting in (2),

– 3y2=21 ;
9

3
... 75+120y+48yo-15-12y2-27y=189;

9y2+1057–114=0,

3y2 +354-38=0; (y-1)(3y+38)=0;

38 .: Y=l, or

3

137 and by substituting in (1),

X=3, or

9

SYMMETRICAL EQUATIONS. 227. The following method of solution may always be used when the given equations are symmetrical, that is, when the unknown quantities in each equation may be interchanged without destroying the equality. The same method may generally be employed with advantage where the given equations are symmetrical except with respect to the signs of the terms. Example. Solve 204 + y4=82.

(1), 3 Y=2

.(2). Put

x=u+v, and y=u-V; then from (2) we obtain

v=1. Substituting in (1), (u+1)4+(u-1)4=82;

.. 2(4+6u2+1)=82;

44 +6u2–40=0; whence

u2=4, or – 10; and

u=+2, or +V-10. Thus,

=u+v=3, -1, 1+ V-10;

y=u-v=1, -3, -12V-10. Note. We may assume x+y=2u and x-y=2v, u and v being any unknown quantities, whence we obtain x=u+v, and y=u-v, the values used in the above.

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