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From (1.), squaring, 2 - 2 acʻy2 + 4 = 25, and from (2.), squaring, &c., 4 x*y* = 144 Then, adding,

+ + 2 xy2 + y4 = 169 and taking the square root, c2 + y2 = + = 13....... (3). (3.) + (1.), then, 2 x = 18 or – 8, or a = 9 or – 4,

and .. x = † 3 or £ 2V - 1. (3.) - (1.), then, 2 y = 8 or – 18, or yo = 4 or – 9,

and :: y = + 2 or + 3N - 1. NOTE.— The student will see that the pairs of values which satisfy the given equations are, x = 3, y = 2; x = - 3, y = -2; x = 2 N - 1, y = 3 V-1; x = - 2V - 1, y = - 3V - 1. Ex. 3. Solve ac + y = 11 .................................(1.) .

ya + x = 11 y........... ............... (2.) ) Subtracting, then, sc – yo — x + y = 11 x – 11y;

or, 2" = 12 (oc y). Now (20 — y) is a factor of each side, and hence, striking it out, we have

2 + y = 12 ..........................................(3.), and also x - y = 0....

............(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.), x2 – 2 = 11 x – 12,

from which w = 6 + 2/6; and hence from (3.), by substitution, we easily get

y = 6 7 2 16. Again, taking equations (4.) and (1.)- .

we have, from (4.) x = y, and :: from (1.), aca + x = 11 x or x2 = 10 x,

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

are

Hence, the pairs of values satisfying the given equations

x = 10, y = 10; x = 0, y = 0;
a = 6 + 2 N/6, y = 6 - 2 /6 ;

c = 6 - 2 N/6, g = 6 + 2 6. NOTE.-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 x = 0, y = 0 will always satisfy. Ex, 4. Solve 3 ac – 2 xy = 55 ........ ....... (1.) 1

da – 5 xy + 8 ya = 7 ............... Multiplying the equations together crosswise, we get

55 22 – 275 xy + 440 ya = 21 oca – 14 xy; or, transposing, 34 24 – 261 xy + 440 y = 0;

or, (2x - 5y) (17 aC – 88 y) = 0,

from which 2 x = 5 y, and 17 x = 88 y. Each of these equations taken in turn with either (1.) or (2) will easily give the required values of x and y. Ex. 5. a* + y4 = 337...

.............. (1.), C + y = 7..

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

204 + 4 xoy + 6 x+ya + 4 xy + y4 = 2401 ;..........(3.) (3) - (1), then 4 x®y + 6 xʻrja + 4 xyl = 2064;

or, 2 x®y + 3 x*ya + 2 xy: = 1032; or, arranging, 2 xy(x + y) – ay* = 1032

but from (2), (oc + y)= 49, and hence, 2 xy(49) – dʻya = 1032 ; or, woya – 98 xy + 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 ac and y.

Ex. III. 1. x + y = 5, xy = 6. 2. a - y = 2, xy = 15. 3. 2 + y2 = 25, xy = 12. 4. co + y2 = 20, x + y = 6.

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ya = 29, x - y = 3.

ya = 13, (2 – y)2 = 1. 7. oc – y2 = 27, xy = 18.

ya = 12, + y = 6. + y = 53, ** – yo = - 45. 2c2 + xy = 28, ya + xy = 21. 11. 2c + xy + y2 = 19, xy * = - 3. 12. x + y = 13, Nac + Ny = 5. 13. ^ + y + y^ = 84, a + Mg + y = 14. 14. 32 + y2 = 35, xʻy + xy = 30. 1 1 1 1 1

+ = = d, t = 0. ac + y = a(x y), a2 + y = lo. 17. 004 y = a, ac ya = b. 18. R + y = 5, 2 + y = 35. 19. x + y = 5, 2x + yo = 275. 20. x2 + y2 = ¥ (x + y), xy = 6. 21. x - y = 2, 2083 – = 98. 22. xy(x + y) = a, * y* (x + y) = b. 23. xy(x + y) = 30, x*y* (206 + y) = 9900. 24. 4 x2 – 3 xy = 18, 5 ye 2 xy = 8.

30 25. a* + y = 35, až + y = 1

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26. m2 + yš = 3 x, a1 + y} = x.

27. xy + 6 = 226, x + y + 4 =

28. (c + y)2 + 2(x - y)2 = 3 (+ y) (cc – y), c* + y = 10. 29. x2 + 10 xy + y2 = (22 – y?), x2 + 5y = x + 13 y. 30. 2* – 2 x+ya + 24 = 1 + 4 xy, ac* (20 + 1) + y(y + 1) = xy.

9 202 72 - 117 8 x2 - y2 + 1 . 31. 3x + y - 9 =

32 – Y

6 4x + y + 1

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20 + 2, Y + 2 X + 2, Y -
y - 2*– 2

Y + 2.
Vaca - y2 + (x - y) ,

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Vaca - j - (« – y)
34. y - 2 15 22 + y + 3 = 32 – 5 x®,

6 +3* + 6 + 2) = 14%. 35. – - - 3 = - Y, 3 x + y = 7.

Y 2 36. (c + y) ay = c (bac + ay), cy (ba + ag) = + abc (a + y - c). 37. Y* = *(ay bx), aä = ax by. 38 Na + 1 + 1 Vic + 9 + 3 Y

või

= ,x(y + 1)= 36(33 + 10). 39. x = a... 6 Y +

+ Ź 2 + y 40. x + y + z = 6, xy + x2 + yz = 11, xyz = 6. 41. aa yz = 0, y2 – 2z = 0, 2 xy = 0. 42. xyz = a'(c + y) = 6?(y + x) = c+(20 + x). 43. x2 + y2 + x2 =

2 a 32 ca

44. C + y + 2 + u = 4 a + 4b,

ay + 2% + xu + yz + yu + zu = 6 a® + 12 ab +66, xyz + xyu + czy + yzu = 40% + 12 a+b + 12 ab2 + 46,

xyzu = a* + 4 a3b + 6 a2b2 + 4 ab3 + 64. 45. xʻy2 + xyʻz + acʻyz = a,

yax2 + xyʻ% + xyza = b,

2 mm + acʻyz + xyz = C. 46. (ac + y)2 + m2 = 1125, C + y + z = 15, 2y = 24.

1 1 1 47. If ax3 = by3 = cm), and + + = a, show that ax" + bya + cze = (+ b3 + ct )3 a?.

48. Given R = 1 + %, P = 4 (1 R-4), M = PR", show that R = 4(1 - 1) +1.

then

100 +

CHAPTER II. Problems Producing Quadratic Equations. 17. 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 31 less per cent. than he had raised them. Find the original rise per cent. Let u = the original rise per cent.,

8C

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

100 C
2 – 100 + 7 = 3}, which solved, gives

= 20 or - 16%. The value a = 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, 2 = – 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 31 more per cent. than he had lowered them. Find the original fall per cent.

The above solution tells us that the fall required is 16; 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.

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