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From (1.), squaring, 2C+ 2 x*y + y 25, and from (2.), squaring, &c., 4 a'y

144 Then, adding,

20* + 2 x*ya + y4

169 and taking the square root,

x2 + y = + = 13.......(3). (3.) + (1.), then, 2 x? = 18 or

8, or x

- 9 or – 4, and ... 30 =

+ 3 or + 2 = 1. (3.) - ) (1.), then, 2 y = 8 or - 18, or yo = 4 or - 9,

and :: y = $ 2 or + 3 = 1. NOTE.—The student will see that the pairs of values which satisfy the given equations are, w = 3, y = 2; x = 3, y = -2; x = 2N - 1, y = 3 V-1; x =

- 2V - 1, y =

- 3V - 1. Ex. 3. Solve ac + Y

.

(1.) ya + 3 =

(2.) S Subtracting, then, x - y - x + y = 11x - 11 y;

or,

aca ya 12 (oc y). Now (x - y) is a factor of each side, and hence, striking it out, we have x + y

(3.), and also XC y

.(4.). Equations (3.) and (4.) may not be used as simultaneous equations, but each of them may be used in turn with either of the given equations. Thus, taking equations (3.) and (1.), we have— (1.) – (3.), 22 - x = 11x

12, from which x = 6 + 2 V6; and hence from (3.), by substitution, we easily get

y = 6 3 2 J. Again, taking equations (4.) and (1.)

we have, from (4.) x = y,
and :: from (1.), ac + 3 = 11 w or ac =

12...

= 0..

from which x = 10 or 0; and so, from (4.), y 10 or 0.

10 x,

Hence, the pairs of values satisfying the given equations

are

2 wy

=

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14 wy;

88 y.

10, y = 10; x = 0, y = 0; 2 = 6 + 2 /6, y = 6 – 2/6 ;

c = 6 - 2 6, y = 6 + 2 /6. NOTE.-It worth while remarking that when each of the terms of the given equations contain at least one of the unknown quantities, the values 3 = 0, y = 0 will always satisfy. Ex. 4. Solve 3 XC

55......

(1.) acea 5 xy + 8y = 7

(2.) Multiplying the equations together crosswise, we get

55 m2 275 xy + 440 ya 21 cm or, transposing, 34 2* - 261 xy + 440 y = 0;

or, (2 x 5 y) (17 2 88 y) = 0, from which 2x = 5

Y,

and 17 * = Each of these equations taken in turn with either (1.) or (2) will easily give the required values of x and y. Ex. 5. 204 + y4

(1.), 2 + y 7

(2.) From (2), raising each side to the fourth power, we have + + 4xy + 6 x+y + 4 xy + y = 2401 ;..

2401 ;......... (3.) (3) - (1), then 4 xy + 6 x*y2 + 4xyz

or, 2xy + 3a'y + 2 xys = 1032;
or, arranging, 2 xy(x + y)2 - x*y* = 1032 ;

but from (2), (x + y)?
and hence, 2 xy(49) – *y= 1032;
xʻy

0, from which xy 12 or 86,......(4.) From (2) and (4), x - y may now be easily obtained, and hence also the required values of w and y.

337...

2064;

49,

or,

983

sy + 1032 =

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27, wy

= 18.

53, wcz

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- 45.

19, wy

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+

+ X

13.

а, а?

38 = 35.

5. 2c + y2 29, 3

y = 3. 6. Beca у

13, (OC y) 1. 7. x2 Y 8. x2 y2 12, x + y

6. 9. 2 + YM

y2 10. 2a + xy 28, ya + xy

21. 11. z + xy + ýja

we 3. 12. 2 + y

13, Voc + Vy = 5. 13. u 2 + xy + y2 84, a + Næy + y = 14. 14. 202 + y 35, xy + xy = 30.

1 1 1 1 15.

6.

x2 16. 2 + y = a(oc y), 2 + y2 17. 24 - y

ya b. 18. x + y

5, 20+ 19. 3 + Y

5, 200 + you 275. 20. acé + y = ¥ (2 + y), xy 6.

Y

2, 23 у 98.
22. ay(x + y) = a, a* y*(c* + y) = b.
23. xy(x + y) = 30, x*y (200 + y) = 9900.
24. 4 m2

3 xy
18, 572

8.

30 25. x + y$ = 35, ** + yt

ały?' 26. 2 + yt = 3x,

act + y} 216

77 27. xy + 6

x + y + 4 xy

2 + 3 28. (c + y)” + 2(x y)2 =

+ 2(x - y) = 3 (x + y) (? y), a* + y* = 10. 29. 22 + 10 xy + ya = + (2* - y), ** + 5y = x + 13 y. 30. ach - 2 cʻy

2 xʻya + y 1 + 4xy, 2c*(x + 1) + y(y + 1)

21. aC

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= *.

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35.

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bx), a

= ax

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y

wy
3

3 x + y = 7.
y

aca 36. (2 + y) xy = c (6x + ay), my (bac + ay) cy + abc (z + 9 - c). 37. y4 2+(ay

by. Nya + 1 + 1 Nã

3 + 9 + 3 38.

(y + 1)2 = 36(48 + 48). y

6 39. X =

y
x + 7

20 + Y 40. 3 + y + % = 6, xy + x2 + y2

6. 41. ca

yz
1, ya

ху

0.
a(+ y) 8(y + 2) c"(x + 2).

72 43. ma + y + 32

с

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11, aya

C2 =

42. xyz =

44. x + y + % + u = 4 a + 4b,
xy + x2 + xu + y2 + yu + zu =

6 a2 + 12 ab +66%, xyz + xyu + Xzu + YZU 4a3 + 12 a+b + 12 ab? + 465, XYZU

at +. 4 a36 + 6 a2b2 + 4 ab3 + 64. 45. Qoy + xyz + x*yz = a,

y** + xyʻz + wyze = b,

dama + acʻyz + xyz c. 46. (a + y) + 28 = 1125, 2 + y + z = 15,

ху
24.

1 1 1 47. If anca by and +

= a, show that

y ax2 + by2 + c2 (a} + b3 + ct ) a?.

CA,

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.

7. We shall now discuss one or two problems whose solutions depend upon quadratic equations.

Ex. 1. A person raised his goods a certain rate per cent., and found that to bring them back to the original price he must lower them 3} less per cent. than he had raised them. Find the original rise per cent.

Let x = the original rise per cent., then

100 = the fall per cent. to bring them to 100 + 2C the original price. Hence, by problem

100 aC

34, which solved, gives 100 + 2C

16. The value x = 20 is alone applicable to the problem. Remembering, however, the algebraical meaning of the negative sign, it is easy to see that the second value, a 163, gives us the solution of the following problem:

A person lowered his goods a certain rate per cent., and found that to bring them back to the original price he must raise them 3į more per cent. than he had lowered them. Find the original fall per cent.

The above solution tells us that the fall required is 163

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X =

20 or

per cent.

Had we worked the latter problem first, we should have obtained x = 16 or 20, the value o = - 20 indicating the solution of the former problem.

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