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y + 2
x + 2
2 + 2
1 1 2 33.
= 3 + 2 12, √x - y (oC y)
3 x + y = 7. y
2 - y2 36. (c + y) xy = c (bxc + ay), xy (bxc + ay) 2ʻya + abc (3C + y - c). 37. y* = x*(ay – bx), xa by.
Ny + 1 + 1 √x + 9 + 3 38.
a(y + 1)2 = 36(43 + 40). y
6 39. C = y +
2 + y 40. 3C + + 3 == 6, xy + 2% + yz 11, xyz = 6. 41. Oca y% 0, y2 0, – Xy
0. aʼ(2 + y) bly + x) c'(x + ).
ao 72 ca 43. a2 + y2 + 22
ay + x2 + xu + y2 + yu + zu = 6 a2 + 12 ab +
40% + 12 a+b + 12 ab+ 46, Xyzu
a* + 4 a3b + 6 aoba + 4 ab3 + 64. 45. a”ya + xyʻz + x^yz = a, yama + wyʻz + xyza
b, vana + x^yz + xyza 46. (x + y) + 2,3 = 1125, C + + x
1 1 1 47. If ax3 bys cm), and
a, show that
Y axa + bya + cza = (aš + b3 + ct ) a.
42. wym =
100 + 2C
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 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 the original price. Hence, by problem
3}, which solved, gives 100 + OC
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 34 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
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.
Let x =
Ex. 2. Find a number such that when multiplied by its deficiency from 100 the product is 196.
the number, then 100 x = its deficiency from 100. Hence, by problem
2 (100 – æ) 196, or ac 100 x + 196 from which, x = 2 or 98.
Both these values will be found to be consistent with the conditions of the problem.
Ex. 3. The number of men required to build a house is such that, when four times the number is subtracted from three times the square of the number, the result is 160. Find the number of men.
the number of men, then, by problem
160, from which
6. The value o = 8 is alone applicable to the problem as it stands. If, however, we may conceive of a fractional number of men—and this we may easily do here by supposing a boy's work to be equal to of a man's—we find that the second result gives us the solution of the following problem:
The number of men required to build a house is such that when four times the number is added to three times the square of the number, the result is 160. Find the number.
The answer, as above indicated, is 6 men and I boy, where a boy is worth of a man.
The student will find, however, that in some cases there is no obvious interpretation to the second result, owing occasionally to the fact that certain terms are used in the problem to which the results will not apply, and indeed that the algebraical expression of the conditions of the problem is more general than the language of the problem itself.
Ex. IV. 1. Find a number whose square is equal to the product of two other numbers, one of which is less by 6 than the required number, and the other greater by 9 than twice that number,
2. When the numerator and denominator of a certain fraction are each increased by unity the fraction is increased by ito, and when they are each diminished by unity the fraction is diminished by o's.
Find the fraction. 3. The mean proportional between the excess of a certain number above 21, and its defect from 37, is 28. Find the number,
4. A number of articles, which were bought for £4, cost each 3 shillings more than half as many shillings as there were articles. Find the number of articles,
5. There is a square court-yard, such that if its length be increased by 10 feet, and its breadth diminished by 20 feet, its area would be 5,104 feet. Required the side of the square.
6. If the number of shillings given for an article be added to the number of articles which can be bought at the same price for 18 shillings, the result is 11. Find the price.
7. Two travellers set out to meet each other from two places 180 miles distant; the first goes 3 miles an hour, and the second goes 1 mile more per hour than one-fourth of the number of hours before they meet. When will they meet?
8. A farmer bought a number of calves, sheep, and pigs, the number of calves being equal to that of the sheep and pigs together. For the calves he gave 64s. a head, and for the sheep twice as many shillings as there were sheep. He paid £153. 12s. for the calves and sheep together, and £36. 12s. for the pigs—a pig costing as much as a sheep and calf together. Find the cost of the sheep per head.
9. There are two squares, and an oblong whose sides are equal to those of the squares, and it is noted that three times the area of the first square exceeds four times the area of the oblong by 3 square feet, while twice the area of the
square, together with three times the area of the rectangle, is 36 square feet. Required the sides of the squares.
10. The sum of two quantities is equal to 6 times the square of their product, and the sum of their cubes is equal to 36 times the product of their fifth powers. Find the quantities.
11. The solid content of a rectangular parallelopiped is 60 cubic feet, and the total area of the side is 98 square feet, while the sum of the edges is 48 feet. Required the dimensions.
12. The products of the number of units of length in the sides of a polygon of n sides, when taken n - 1, together are respectively a, a,, az, &c., a Required the lengths of the sides.
13. A, B, and C can together do a piece of work in a day, and C's rates of work is the product of the rates of A and B. Moreover, C is one-fifth as good a workman as A and B together. Find the respective times required for A, B, C to do a piece of work.
14. The compound interest of a certain sum of money for 3
years is a, and the third year's interest is b. Find the principal and the rate per cent.
15. A owes B £a due m months hence, and also £b due n months hence. Find the equated time, reckoning interest at 5 per cent. per annum.
16. Find three quantities such that the sum of any two is equal to the reciprocal of the third.
17. Find three magnitudes, when the quotients arising from dividing the products of every two by the other are respectively a, b, c.
18. The sum of three quantities is 9, the sum of their products, taken two and two together, is 23, and their continued product is 15. Show that the three quantities are the roots of the equation as 9 a + 23 a 15 = 0.
8. We shall reserve the discussion of the complete theory of Indices for Vol. II., confining ourselves here to a few simple cases, and giving a few examples involving fractional and negative exponents.
9. Fractional exponents.
DEF.—The numerator of a fractional exponent indicates the power to which the quantity must be raised, and the