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228. The examples we have given will be sufficient as a general explanation of the methods to be employed; but in some cases special artifices are necessary. Example 1. Solve x2y2 – 6x=34-3y

-(1), 3xy+y=2(9+)

(2). From (1)

2c2y2 – 6x+3y=34 ; from (2)

9xy 6x+3y=54; by subtraction,

x+y2 – 9xy=-20, 9+V81-80

9+1 Xy=

:=5 or 4.
2

2
(i) Substituting xy=5 in (2) gives y-2x=3.

5
From these equations we obtain x=1, or

2'

y=5, or -2. (ii) Substituting xy=4 in (2) gives y-2x=6.

-3+ 17 From these equations we obtain x=

2 and

y=3+ 17. Example 2. Solve 32 + y2 +22=49

(1), z2+zx+x2=19

(2), +y+y2=39 Subtracting (2) from (1)

y2-3°+2(y-2)=30; that is, (y-2)(x+y+z) =30

(4). Similarly from (1) and (3)

(2-3)(x+y+z)=10. Hence from (4) and (5), by division

Y-3

=3; whence

y=32– 2x. Substituting in equation (3), we obtain

x2 – 3x2+322=13. From (2),

22+xz+22=19.
Solving these homogeneous equations, we obtain

x=+2, z=+3; and therefore y=+5;
11
1

19
x=

25+ ; and therefore y=F

(3).

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sy=10.

EXAMPLES XXVIII. C. 1. 5x Y=17, 2. 2x2 + xy=15, 3.

3-y=10, xy=12. y2 + xy=10.

22 - 2xy 3y2=84. 4. 3x+2y=16, 5. 3.c-y=11, 6.

x-3y=1, 3x2 - y2=47.

x2 – 2xy +9y2=17. +2y=9, 8. 22 + y2=5, 9.

5x+y=3, 3y2–5x2=43. 2xy-y2=3.

2x2 – 3xy-y2=1. 10. 3x2 — 5y2=28, 11. 3x2 — y2=23, 12. 22 + xy + y^=3}, 3xy-4y2=8. 2x2 — xy=12.

2.2-3xy+2y=23. 13. 2-3.cy+y2+1=0,

14. 7xy - 8x2=10, 3.2 - xy+3y=13.

8y2 — 9xy=18. 15. 22 — 2xy=21, 16. x2 + 3xy=54, 17. 23+y3 =152, xy + y2=18.

xy + 4y2=115. r4y+xy2=120. 18.

23 — Y3=127, 19. 23 y3=208, 20. x?y2 + 5xy=84, 2-y xy2=42. xy(3-y)=48.

3+y=8. 21. x2 +4y2 +80=15x+304, 22. 9x2 + y2 – 63x — 217 +128=0, XY=6.

XY=4.
1 1
45

1 1 243
y2 4'

+

203 y 8 1 1 3

1 1 9

+
y
2

Y 25. 3+++ya+y=931,

26. x2 + xy + y2=84, 82 - xy + y2=19.

Vcg +y=6. 27. + xy+y=65,

28. x+y=7+ vay, x2 + xy + y2=2275.

2x2 + y2=133-wy. 29. 3x2-5y4=7, 30. 5y2 - 7x2=17, 31. 3x2 +165=16xy, 3.y - 4y=2. 5xy 6x2=6.

7xy+3y2=132. 32. 3x2 + xy+y2=15,

33. 2x2 + y2 —3=3xy, 31xy - 3x2 — 5y2=45.

2x2–6+y2=0. 34. 204 + y4=706, 35. 204 + y4=272, 36. 205 y5=992, Katy=8. X-Y=2.

3 - Y=2. 37. Byt -6ya= -9,

38. 2008 + 2y8=9xy, Xy-Y=2.

x+y=3. 39. x+y=1072, 40. Qy}+yxt=20, 41. 2*+yš=5, act+y=16.

22+y=65. 6(x-+y-b)=5.

23.

+

24:

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43.

42. Vc+y+ V2-y=4,

y+ V3c2—1=2, 22 —-y2=9.

#+1-12-1= vy. 10 44. +

45. VI-VY, VI+ vy_17 Vy 3'

+
vx+vy VX-VY

4 3+y=10.

32+y=706. 46. 4x2 +5y=6+20xy-25y2 + 2x, 47. Py2 +400=41cy, 7. – 1ly=17.

y2 +4x2=5xy. 48. 9x2 +33x – 12=12xy - 4y2 +22y, 49. xy+ab=2ax, x? — Xy=18.

c?y2 +a%b2=2boy (x2 - y2)(x-y)=16xy,

51. 2x2 — xy + y2=2y, (x-7)(x2 - y) =640x4y.

2x2 + 4xy=5y. 22 + y2 -2=21,

53. 2-Y-2=2, 3xz+3yz2xy=18,

22+y2-22=22, 2+y-2=5.

xy=5. 54. 2+yz+22=49,

22+z+2=19,
2c2 + xy + y2=39.

50.

52.

CHAPTER XXIX.

PROBLEMS LEADING TO QUADRATIC EQUATIONS.

C

or

229. We shall now discuss some problems which give rise to quadratic equations.

Example 1. A train travels 300 miles at & uniform rate; if the rate had been 5 miles an hour more, the journey would have taken two hours less : find the rate of the train. Suppose the train travels at the rate of x miles per hour, then the

300 time occupied is hours.

300
On the other supposition the time is hours;

*+5
300 300
- 2

(1): * +5 whence

2 + 5x – 750=0, (x +30) (2 – 25)=0,

.. X=25, or – 30. Hence the train travels 25 miles per hour, the negative value being inadmissible.

It will frequently happen that the algebraical statement of the question leads to a result which does not apply to the actual problem we are discussing. But such results can sometimes be explained by a suitable modification of the conditions of the question. In the present case we may explain the negative solution as follows.

Since the values x=25 and - 30 satisfy the equation (1), if we write - x for x the resulting equation,

300 300
-2.....

.(2), - 2 + 5 will be satisfied by the values x= -25 and 30. Now, by changing

300 300 signs throughout, equation (2) becomes

+2;

5 and this is the algebraical statement of the following question:

A train travels 300 miles at a uniform rate; if the rate had been 5 miles an hour less, the journey would have taken two hours more: find the rate of the train. The rate is 30 miles an hour.

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that is,

Example 2. A person selling a horse for $72 finds that his loss per cent. is one-eighth of the number of dollars that he paid for the horse : what was the cost price ?

Suppose that the cost price of the horse is x dollars ; then the loss on $100 is $2. 8

x2 Hence the loss on $x is x x

or

dollars; 800 800

22 .: the selling price is x—

dollars. 800

. ? Hence

= 72,

800
x2 – 800x + 57600=0;
(x – 80) (x – 720)=0;

= 80, or 720; and each of these values will be found to satisfy the conditions of the problem. Thus the cost is either $80, or $720.

Example 3. A cistern can be filled by two pipes in 33} minutes; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly.

Suppose that the two pipes running singly would fill the cistern in & and 2-15 minutes. When running together they will fill 1

1

3 of the cistern in one minute. But they fill -- 15

331' 100 of the cistern in one minute;

1

3 Hence

100'
100 (2x – 15)= 3x (x – 15),
3.x2- 245x + 1500=0);
(oc – 75) (3x – 20)=0;

. . x=75, or 6%. Thus the smaller pipe takes 75 minutes, the larger 60 minutes. The other solution 63 is inadmissible.

Example 4. The small wheel of a bicycle makes 135 revolutions more than the large wheel in a distance of 260 yards; if the circumference of each were one foot more, the small wheel would make 27 revolutions more than the large wheel in a distance of 70 yards; find the circumference of each wheel,

+

or

1

+

2 - 15

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