2d. This negative value, taken with its proper sign, will also satisfy the conditions of the problem, understood in its algebraic

sense.

3d. If a positive result is interpreted in a certain sense, a nega tive result must be interpreted in a directly contrary sense.

4th. The negative result, with its sign changed, may be regarded us the answer to a problem of which the enunciation only differs from the one proposed in this: that certain quantities which were additive have become subtractive, and the reverse.

90. As a further illustration of the extent and power of the algebraic language, let us resume the general problem of the laborer, already considered.

Under the supposition that the laborer receives a sum c, we have the equations

If, at the end of the time, the laborer, instead of receiving a sum c, owed for his board a sum equal to c, then, by would be greater than ax, and under this supposition, we should have the equations,

==

x+y=n, and axby c. Now, since the last two equations differ from the preceding two given equations only in the sign of c, if we change the sign of c, in the values of x and y, found from these equations, the results will be the values of x and y, in the last equa tions: this gives

bn
a + b'

с

The results, for both enunciations, may be comprehended in the same formulas, by writing

The double sign, is read plus or minus, and, is read, minus or plus. The upper signs correspond to the case in which the laborer received, and the lower signs, to the case in

which he owed a sum C. These formulas also comprehend the case in which, in a settlement between the laborer and his

employer, their accounts balance. This supposes c = 0, which gives

91. The discussion of a problem consists in making every possible supposition upon the arbitrary quantities which enter the equation of the problem, and interpreting ne results.

An arbitrary quantity, is one to which we may assign a value, at pleasure.

In every general problem there is always one or more arbi trary quantities, and it is by assigning particular values to these that we get the particular cases of the general problem.

The discussion of the following problem presents nearly all the circumstances which are met with in problems giving rise to equations of the first degree.

PROBLEM OF THE COURIERS.

Two couriers are traveling along the same right line and in the same direction from R' toward R. The number of miles traveled by one of them per hour is expressed by m, and the number of miles traveled by the other per hour, is expressed by n. Now, at a given time, say 12 o'clock, the distance between them is equal to a number of miles expressed by a: required the time when they are together.

At 12 o'clock, suppose the forward courier to be at B, the other at A, and R or R' to be the point at which they are together.

Let a denote the distance AB, between the couriers at 12 o'clock, and suppose that distances measured to the right, from A, are regarded as positive quantities.

Let t denote the number of hours from 12 o'clock to the time when they are together.

Let x denote the distance traveled by the forward courier

in t hours;

Then, a + x will denote the distance traveled by the other in the same time.

Now, since the rate per hour, multiplied by the number of hours, gives the distance passed over by each, we have,

Subtracting the second equation from the first, member from member, we have,

We will now discuss the value of t; a, m and n, being arbitrary quantities.

First, let us suppose m>n.

The denominator in the value of t, is then positive, and since a is a positive quantity, the value of t is also positive.

This result is interpreted as indicating that the time when they are together is after 12 o'clock.

The conditions of the problem confirm this interpretation. For if m>n, the courier from A will travel faster than the courier from B, and will therefore be continually gaining on him the interval which separates them will diminish more and more, until it becomes 0, and then the couriers will be found upon the same point of the line.

:

In this case, the time t, which elapses, must be added to 12 o'clock, to obtain the time when they are together.

Second, suppose m <n.

The denominator, mn will then be negative, and the value of t will also be negative.

This result is interpreted in a sense exactly contrary to the interpretation of the positive result; that is, it indicates that the time of their being together was previous to 12 o'clock.

confirmed by considering the For, under the second suppoadvance travels the fastest, and

This interpretation is also circumstances of the problem. sition, the courier which is in therefore will continue to separate himself from the other courier. At 12 o'clock the distance between them was equal to a: after 12 o'clock it is greater than a; and as the rate of travel has not been changed, it follows that previous to 12 o'clock the distance must have been less than a. At a certain hour, therefore, before 12, the distance between them must have been equal to nothing, or the couriers were together at some point R'. The precise hour is found by subtracting the value of t from 12 o'clock.

This result indicates that the length of time that must elapse before they are together is greater than any assignable time, or in other words, that they will never be together.

This interpretation is also confirmed by the conditions of the problem.

For, at 12 o'clock they are separated by a distance a, and if mn they must travel at the same rate, and we see, at once, that whatever time we allow, they can never come together; hence, the time that must elapse is infinite.

Fourth, suppose a =

O and m>n or m <n.

The numerator being 0, the value of the fraction is 0 or t = 0.

This result indicates that they are together at 12 o'clock, or that there is no time to be added to or subtracted from 12 o'clock.

The conditions of the problem confirm this interpretation. Because, if a = 0, the couriers are together at 12 o'clock; and since they travel at different rates, they could never have been together, nor can they be together after 12 o'clock: hence, t can have no other value than 0.

Fifth, suppose a = 0 and m=n.

0

0'

The value of t becomes an indeterminate result.

This indicates that t may have any value whatever, or in other words, that the couriers are together at any time either before or after 12 o'clock: and this too is evident from the cir cumstances of the problem.

=

For, if a 0, the couriers are together at 12 o'clock; and since they travel at the same rate, they will always be together; hence, t ought to be indeterminate.

The distances traveled by the couriers in the time are, respectively,

both of which will be plus when m>n, both minus when m < n, and infinite when m = n.

In the first case t is positive; in the second, negative; and in the third, infinite.

When the couriers are together before 12 o'clock, the distances are negative, as they should be, since we have agreed to call distances estimated to the right positive, and from the rule for interpreting negative results, distances to the left ought to be regarded as negative.

Of Inequalities.

92. An inequality is the expression of two unequal quantities connected by the sign of inequality.

Thus, ab is an inequality, expressing that the quantity a is greater than the quantity b.

The part on the left of the sign of inequality is called the first member, that on the right the second member.

The operations which may be performed upon equations, may in general be performed upon inequalities; but there are, never theless, some exceptions.

In order to be clearly understood, we will give examples of the different transformations to which inequalities may be sub