Hence, the second courier will overtake the first in 42 hours, and the whole

Let a courier, who travels at the rate of m miles in t hours, be despatched from B in the direction C; and ʼn hours after his departure, let a second courier, who travels at the rate of m' miles in t' hours, be sent from A, which is distant d miles from B, in order to overtake the first. In what time will he

come up with him, and what will be the whole distance travelled by each? Let x number of hours that the second courier travels.

m

t

Then, since the first courier travels at the rate of m miles in t hours, that is,

miles in 1 hour, he will travel x miles in a hours, and since he started t

n hours before the second courier, the whole distance travelled by him will

be (n+x)”.

Again, since the second courier travels at the rate of m' miles in t hours, that

m'

is,

m'

miles in 1 hour, he will travel miles in a hours; but since he started from A, which is distant d miles from B, the whole distance travelled by the

x will be greater than the whole distance travelled by

A father, who has three children, bequeaths his property by will in the following manner: To the eldest son he leaves a sum a, together with the na part of what remains; to the second he leaves a sum 2 a, together with the n part of what remains after the portion of the eldest and 2 a have been subtracted; to the third he leaves a sum 3 a, together with the n" part of what remains after

the portions of the two other sons and 3 a have been subtracted. The property is found to be entirely disposed of by this arrangement. Required the amount of the property.

If we can, by means of this quantity, find algebraic expressions for the portions of the three sons, we must subtract their sums from the whole property ¤, and putting this remainder = 0 we shall determine the equation of the problem. Let us endeavour to discover these three portions.

Since x represents the whole property of the father, x-a is the remainder after subtracting a; hence,

According to the conditions of the problem, the property is entirely disposed of. Hence, when the sum of the three portions is subtracted from x, the difference must be equal to zero; this gives us the equation

2

n3 x-6 an3 — 3 n2 x + 10 a n2 + 3 n x-5 an-x+α = 0 •. (n3—3n'+3n—1) x =6an3-10an2+5an-a

=

=

(6n-10n'+5n-1) a (n-1)3

By reflecting upon the conditions of the problem, we may obtain an equation much more simple than the preceding. It is stated that the portion of the third son is 3 a, together with the n" of what remains, and that the property is thus

entirely disposed of; in other words, the portion of the third son is 3 a, ani the remainder just mentioned is nothing.

We found the expression for that remainder to be

This result is, moreover, more simple than the former.

- 1) a, and (n3.

We can easily prove that the two expressions are numerically identical, for applying to the two polynomials (6 n3 10 n 2 + 5 n — 2 3 n2 + 3 n + 1), the process for finding the greatest common measure, we shall find that these two expressions have a common factor n- -1; dividing, therefore, both terms of the first result by this common factor, we arrive at the second.

The above problem will point out to the student the importance of examining with great attention the enunciation of any proposed question, in order to discover those circumstances which may tend to facilitate the solution; he will otherwise run the risk of arriving at results more complicated than the nature of the case demands.

The above problem admits of a solution less direct, but more simple and elegant than those given above. It is founded on the observation, that after having subtracted 3 a from the former portions, nothing ought to remain.

Let us represent by 71, 72, 73, the three remainders mentioned in the enunciation; the algebraic expressions for the three portions must be,

1o. By the conditions of the problem we have rз = 0.

Hence the third portion is 3 a.

2o. The remainder, after the second son has received 2a +

3o. The remainder, after the eldest son has received a +

But this remainder forms the portion of the two other sons, hence we have

reducing the whole to a common denominator,

3a (n2—2n+1) + (2 a n + a) (n − 1) + a n 2 + 3 ana

2

n2 — 2 n + 1,

-

This solution is more complete than the former, for we obtain at the sam time the property of the father and the expressions for the portions of his thr

sons.

We shall now solve one or two problems, in which it is either necessary or con venient to employ more than one unknown quantity.

Problem 11.

Required two numbers, whose sum is 70 and whose difference is 16,

Let x and y be the two numbers; then, by the conditions of the problem,

which are the two equations required for its solution.

.(1)

.(2)

A person has two kinds of gold coin, 7 of the larger together with 12 of the smaller make 288 shillings; and 12 of the larger together with 7 of the smaller make 358 shillings. Required the value of each kind of coin.

Let x be the value of the larger coin expressed in shillings, y that of the smaller.

Then, by the conditions of the problem,

7x+12 y = 288.........

And,

12 x + 7y = 358....

Multiplying equation (1) by 7, and equation (2) by 12, and subtracting the former product from the latter,

..(1)

.(2)

Substituting this value of x in equation (1), it becomes, ... 168 + 12 y = 288

:: y = 10

The larger of the two coins is worth 24 shillings, the smaller 10 shillings.

Problem 13.

An individual possesses a capital of £30,000, for which he receives interest at a certain rate; he owes, however, £20,000, for which he pays interest at a certain rate. The interest he receives exceeds that which he pays by £800. Another individual possesses a capital of £35,000, for which he receives interest at the second of the above rates; he owes, however, £24,000, for which he pays interest at the first of the above rates. The interest which he receives exceeds that which he pays by £310. Required the two rates of interest?

Let x and y denote the two rates of interest for £100.

In order to find the interest of £30,000 at the rate x, we have the proportion,

In like manner to find the interest of £20,000 at the rate of y

But, by the enunciation of the problem, the difference of these two sums is

£800, hence we shall have, for the first equation,