x-y=1.

+ 1,

76. xy+xz+yz=26,

xy (x + y) + yz (y + z) + xx (≈ + x) = 162,
xy (x2 + y3) + yz (y2 + z3) + xz (x2 + ≈2) = 538.

77. x3+ y3 +23 = x2 + y2 + z3 = x + y + z = 1.

347. We shall now solve and discuss some problems which lead to quadratic equations.

A man buys a horse which he sells again for £24; he finds that he thus loses as much per cent. as the horse cost; required the price of the horse.

Let x denote the price in pounds; then he loses x per cent.

and thus his total loss is xx, that is,

х

100

х8

100

;

but this loss is

also x-24; thus

=x-24;

ll we can infer is, that the price was either £60 or £40, these values satisfies all the conditions of the problem.

Divide the number 10 into two parts, such that their ll be 24.

lenote one part, and therefore 10-x the other part;

hough x may have either of two values, yet there node of dividing 10, so that the product of the two 24; one part must be 4 and the other 6.

person bought a certain number of oxen for £80; ght 4 more for the same sum each ox would have find the number of oxen and the price of each.

80

ote the number, then is the price of each; if he

Ꮳ

Only the positive value of x is admissible, and thus the number of oxen is 16, and the price of each ox is £5.

In solving problems, as in the proposed example, results will sometimes be obtained which do not apply to the question actually proposed. The reason appears to be that the algebraical mode of expression is more general than ordinary language, and thus the equation, which is a proper representation of the conditions of the problem, will also apply to other conditions. Experience will convince the student that he will always be able to select the result which belongs to the problem he is solving, and that it will be sometimes possible, by suitable changes in the enunciation of the original problem, to form a new problem, corresponding to any result which was inapplicable to the original problem. Thus in the present case we may propose the following modification of the original problem; a person sold a certain number of oxen for £80; if he had sold 4 fewer for the same sum, the price of each ox would have been £1 more; find the number of oxen and the price of each.

have

Let x represent the number; then by the question we shall

The roots of this quadratic will be found to be 20 and – 16; thus the number 20 which appeared with a negative sign as a result in the former case, and was then inapplicable, is here the admissible result.

350. Find a number such that twice its square increased by three times the number itself may amount to 65.