Now, let x+y=6, and x+y=1056; then by substituting 3 for a, and 1056 for b, in the formulæ of roots, the values of x and y will be found; that is, x=3+1, or 3-19; and y=3F1, or 3F-19. Or, by substituting the above values of a and b in the equation 10az1—20a3z2+2a3=b, it becomes 30% +540z+486=1056; from which the values of z may be found; whence, by substitution, the values of x and y will be determined, as before.

Ex. 11. Given x+4y=14, and y2+4x=2y+11, to find the values of x and y.

Ans. x=-46, or 2; and y=15, or 3.

Ex. 12. Given 2x+3y=118, and 5x2-7y2=4333, to find

Ex. 13. Given x2+4y2=256-4xy, and 3y2x2-39, to find the values of x and y.

Ans. x=6, or 102; and y±5, or 59. Ex. 14. Given "+y"=2a", and xy=c2, to find the values of x and y.

Ex. 15. Given 2+2xy+y2+2x=120-2y, and xy—y2=8, to find the values of x and y.

√5.

Ans. x=6, or 9, or -9F5; and y=4, or 1, or -3±

Ex. 16. Given x2+y-x-y-78, and xy+x+y=39, to find the values of x and y.

Ꮖ

Ans. a 9, or 3; or—6—39; and y=3, or 9, or -6139.

Ex. 18. Given 1-2x2y+y2=49, to find the values of x and x1-2x2y2+y*—x2+y2=20,

and y.

Ans. x=3, or ±√6, or ±√(306√5);

and y=2, or-1, or (13/5).*

* There are four other values, both of x and y, which are all imaginary.

1

Ex. 19. Given x-x2—3—y, and 4—x=y—y2, to find the values of x and y. Ans. x=4, or ; and y=1, or 24.

1

Ex. 20. Given x2+x—4x2=y2+y+2, and xy=y2+3y, to

find the values of x and y.

Ans. x=4, or 1; and y=1, or —2.

Ex. 21. Given x+xy=56, and y+2y=60, to find the

values of x and y.

Ex. 22. Given x

of x and y.

Ꮖ

Ex. 23. Given

values of x and y.

Ꮖ

Ans. x42, or F14;

and y=3/2, or ±10. -y=15, and xy=2y3, to find the values Ans. x =18, or 12; and y=3, or —21 10x+y=3xy, and 9y-9x=18, to find the Ans. x=2, or -3; and y=4, or §.

Ex. 24. Given x+y: x―y :: 13: 5, to find the values of and y2+x=25, § x and y.

Ans. x=9, or -14,; and y=4, or —64. Ex. 25. Given x'y'-7xy2=1710, and xy-y=12, to find the values of x and y.

Ans. x=5, or, or

- 19

176-2; and y=3, or —15, or

Ex. 26. Given xy+xy2=12, and x+xy3=18, to find the values of x and y. Ans. z 2, or 16; and y=2, or 1. Ex. 27. Given x+y+√(x+y)=6, and x2+y2=10, to find the values of x and y.

Ans. 2=3, or 1; or 4-61; and y=1, or 3, or 41-61.

Ex. 28. Given x2+4√(x2+3y+5)=55—3y, and 6x —7y =16, to find the values of x and y.

- 53

-9/5072

7

430

y=2, or

-166±6 √5072

49

Ex. 29. Given x2+2x3y=441-ry, and ry=3+z, to find the values of x and y.

Ex. 30. Given (x+y)2-3y=28+3x, and 2xy+3x=35, to

(x=5, oror-(-255), y=2, or 3, or F(-255.)

Ex. 31. Given x2+3x+y=73-2xy, and y2+3y+x=44,

to find the values of x and y.

x=4, or 16; or-12F/58, Ans. {y=5, or -7; or -11V58.

ง

=

Ex. 32. Given 136-2xy, and x+y=10, to find

the values of x and y.

Ans.

Sx=6, or 4; or 55(-),
y=4, or 6; or 55√(11).

Ex. 33. Given y1-432-12xy, and y=12+2xy, to find the values of x and

y.

Ans. x=2, or 3; and y=6, or (21)+3.

CHAPTER XI.

ON

THE SOLUTION OF PROBLEMS,

PRODUCING QUADRATIC EQUATIONS.

§ I. SOLUTION OF PROBLEMS PRODUCING QUADRATIC EQUA

TIONS, INVOLVING ONLY ONE UNKNOWN QUANTITY.

364. It may be observed, that, in the solution of problems which involve quadratic equations, we sometimes deduce, from the algebraical process, answers which do not correspond with the conditions. The reason seems to be, that the algebraical expression is more general than the common language, and the equation, which is a proper representation of the conditions, will express other conditions, and answer other suppositions.

PROB. 1. A person bought a certain number of oxen for 80 guineas, and if he had bought four more for the same sum, they would have cost a guinea a piece less. Required the number of oxen and price of each.

80
And =
x 16

80

=5 guineas, the price of each.

The negative value (-20) of x, will not answer the condition of the problem.

PROB. 2. There are two numbers whose difference is 9, and their sum multiplied by the greater produces 266. What are those numbers?

Let x= the greater; .. x-y= the less.

..x=14, or -91; and x-9=5, or 181.

Here both values answer the conditions of the problem.

PROB. 3. A set out from C towards D, and travelled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went every day one-nineteenth of the whole journey; and after he had travelled as many days as he went miles in one day, he met A. Required the distance of the places C and D.

Suppose the distance was x miles.

..

19

=

the number of miles B travelled per day; and also

= the number of days he travelled before he met A.

+32+: 361

19

19

8, or 4; and x=152, or 76, both which values answer the conditions of the problem. The distance therefore of C from D was 152, or 76 miles.

PROB. 4. To divide the number 30 into two such parts, that their product may be equal to eight times their difference.

Let x= the lesser part; ... 30—x= the greater part, and 30-x-x, or 30-2x= their difference.

Hence, by the problem, x(30——x)=8(30——2x), or 30x--x2 =240—16x; .'. x2-46x=—240.

... completing the square, x2-46x+529=289; ..x=23±17=40, or 6= lesscr part; and 30—-x=30—6=24=greater part. solution of the equation gives 40 and 6 Now as 40 cannot possibly be a part of 30, we take 6 for the lesser part, which gives 24 for the greater part; and the two numbers, 24 and 6, answer the conditions required.

In this case, the for the lesser part,

PROB. 5. Some bees had alighted upon a tree; at one flight the square root of half of them went away; at another eightninths of them; two bees then remained. How many then alighted on the tree?

Let 2x2 the number of bees; x+

16x2
9

+2=2x2,

;

or 9x+16x+18=18x2 ; 2x2-9x=18 Multiplying by 8, 16x2-72x=144; adding 81 to both sides, 16x2-72x+81=225; 4x=9+15=24, or -6; and x=6, or —11. .. 272, or 41. But the negative value -12 of x, is excluded by the nature of the problem; therefore, 72= number of bees.

365. If, in a problem proposed to be solved, there are two quantities sought, whose sumn, or difference, is equal to a given quantity, for instance, 2a; let half their difference, or half their sum, be denoted by ; then r+a will represent the greater, and a the lesser, (Art 102). According to this method of notation, the calculation will be greatly abridg ed, and the solution of the problem will often be rendered very simple.

PROB. 6. The sum of two numbers is 6, and the sum of their 4th powers is 272. What are the numbers?

Let x=

half the difference of the two numbers; then 3+ x= the greater number, and 3-x= the lesser.

.. by the problem, (3+x)'+ (3--x)'=272, or 162+108.r2+2r=272; from which, by transposition and division, x+54x2=55:

.. completing the square,

+54x2+729=784, and extracting the root, 22+27±28; ..2728, and z=1, or

-55.

Now, by taking the positive value, +1, for x, (since in this case, it is the only value of x which will answer the problem); we shall have 3+1=4= the greater, and 3-1=2= the lesser.