18 or

9 or

8 or

4 or

11 ..

11 y..

11 x

11 y;

or, oca

a

From (1.), squaring,

act 2 x+y + 34 25, and from (2.), squaring, &c., 4 ry

144 Then, adding,

oct + 2 xya + Y* 169 and taking the square root,

13.......(3) (3.) + (1.), then, 2 ca

8, or u

4, and ... X =

† 3 or † 21. (3.) - (1.), then, 2 ya

18, or y

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 = 2 N - 1, y = 3V-1; 2 = 2V - 1, y =

3V - 1. Ex. 3. Solve ac2 + y

(1.) ya + 2 =

(2.)) Subtracting, then, x - y2

x + y =

ya 12 (3C - y) Now (x - y) is a factor of each side, and hence, striking it out, we have— 2C + y = 12 .........

(3.), and also a y

0........ 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.), ac

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

y = 6 7 25.

I
Again, taking equations (4.) and (1.)-

we have, from (4.) x = Y,
and :: from (1.), x2 + x =

2

Hence, the pairs of values satisfying the given equations

are

C =

=

=

C =

14 wy;

88 y.

x = 10, y = 10; x = 0, y = 0;

);
a = 6 + 2 J6, y = 6 - 2 /6;

6
a = 6 - 2/6, y = 6 + 2/6.

6 Nore.— It is worth while remarking that when each of the terms of the given equations contain at least one of the unknown quantities, the values a 0, y = 0 will always satisfy. Ex. 4. Solve 3 x*

55.........

(1.) ? ac - 5 xy + 8 ya = 7

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

55 ca 275 xy + 440 y 21 x or, transposing, 34 24 - 261 xy + 440 yo = 0;

or, (2 - 5y) (17
x

88 y) = 0,
from which 2x 5

Y,

and 17 2 = Each of these equations taken in turn with either (1.) or (2) will easily give the required values of x and y. Ex. 5. 24 + y4 = 337.

.(1.), 7

(2.) From (2), raising each side to the fourth power, we have

3c4

** + 4 xroy + 6 x*y* + 4 xy + y = 2401 ;... ... (3.) (3) - (1), then 4 a*y + 6 x*eja + 4 xy = 2064;

or, 2 y + 3x+y + 2 y
or, arranging, 2 xy(x + y)2 – x*yja = 1032 ;
but from (2), (2 + y)?

49,
and hence, 2 xy(49) – a*yo = 1032;
or, acʻya – 982

ху

1032 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 x and y.

1032;

=

Y = 3.

19, xy

=

+

=b.

A, och

ya = 6.

5. aa + ya 29, 6. nca ya

13, (2 – y)2 = 1. 7. co - g° = 27, cy

18. 8. 2 ya 12, ac + y

6. 9. 202 + zja = 53, 22 ya 45. 10. 2a + 2y = 28, ya + xy

21. 11. 2c + xy + y2

oca 3. 12. 3 + Y

13, VoC + Vy 5. 13. c + xy + ya 84, z + x + y 14. 14. 2 + y = 35, aʻy + xy* go go

| 30.
1 1 1 1
15.

+ ag
Y

ya 16. x + y

a(oc – y), ac + 3a = 62. 17. 24

y 18. 2 + y

5, 202 + yg 35. 19. 3 + Y

5,20 + y = 275. 20. ? + y= ( + y), y = 6.

6. 21. X Y

98. 22. xy(x + y) = a, *Y* (2x2 + y) = b.

x 23. xy(x + y) = 30, 2*y* (2 + y) 9900. 24. 4* – 3 xy = 18,5 yo – 2 xy = 8. 4

8

30 25. c + y= = 35, c + a

ały 26. + yš = 3 x, x1 + ys 216

77 27. xy + 6 x + y + 4 xy

X + Y 28. (c + y)2 + 2(x – y)= 3 (oC + y) (oc – y), ** + y = 10. 29. 2a + 10 xy + y2

” (2c2 ya), aca + 5y2 = x + 13 y. 30. 2* – 2 x+ya + y = 1 + 4 xy, ac* (x + 1) + y*(y + 1)

1) x = sy. 9 * - y - 117 8 x?

ya + 1 31. 3.c + y - 9

2, 303

y

=X.

XY

و - شه .4

bx), ac

= ax

a

с

C +

xy = 0.

42. wym

y 35. 3

3 x + y = 7. y 36. (x + y) xy = c (b.c + ay), wy (bac + ay)

c° + abc (z + 9 - c). 37. Y* x*(ay

by. Na + 1 + 1 Nã

x + 9 + 3 38.

(y + 1)2 = 36(y + 10). ข

doc

6 39. 2 = () ข y + s

2 + y 40. c + y + z = 6, xy + x2 + yz 11, xyz = 6. 41. cca

yz

0, go 0,22
a*(x + y) b?(y + x) = c+(c + ).

ao 72 ca 43. 2 + y2 + 2?

aca ya 44. C + y + % + U 4 a + 4b,

wy + x2 + x3 + ym + yu + zu = = 6 ao + 12 ab +66, xyz + XYU + XZU + yzu 4a® + 12 a+b + 12 ab? + 463, Xyzu

a* + 4 aș5 + 6 aʼba + 4 ab3 + 64. 45. doʻya + xyʻz + acʻyz a, y* * + cy®z + xyz* = b,

b
ana + acʻyz + xyza
46. (c + y)+ 3 = 1125,
C + y + z = 15,

ху
24.

1 1 1 47. If ax? bys cm), and

= a, show that

y axa + bya + cza (aš + b3 + ct ) a?.

= C.

+

+

+

Let 3 =

Problems Producing Quadratic Equations. 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.

the original rise per cent., then 100 + 2C

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

31, which solved, gives 100 + 2C

163 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, . 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 3} 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

100 x

20 or

per cent.

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