1. To find a number, which added to the number 6, will give a sum equal to the number a.

Let x = the required number.

Then, by the conditions

x+ba, whence xa — b.

This expression, or formula, will give the algebraic value of x in all the particular cases of this problem.

This last value of x, is called a negative solution. How is it to be interpreted?

If we consider it as a purely arithmetical result, that is, as arising from a series of operations in which all the quantities are regarded as positive, and in which the terms add and subtract imply, respectively, augmentation and diminution, the problem will obviously be impossible for the last values attributed to a and b; for, the number b is already greater than 24.

Considered, however, algebraically, it is not so; for we have found the value of x to be – 7, and this number added, in the algebraic sense, to 31, gives 24 for the algebraic sum, and therefore satisfies both the equation and enunciation.

2. A father has lived a number a of years, his son a number of years expressed by b. Find in how many years the age of the son will be one fourth the age of the father.

Suppose a 54, and b:

=

= 9; then ax

18

6.

The father being 54 years old, and the son 9, in 6 years the father will be 60 years old, and his son 15; now 15 is the fourth of 60; hence, x = 6 satisfies the enunciation.

If we substitute this value of x in the equation of condition,

Hence,

10 = 10.

5 substituted for x, verifies the equation, and therefore is a true answer.

Now, the positive result which was obtained, shows that the age of the father will be four times that of the son at the expiration of 6 years from the time when their ages were considered; while the negative result, indicates that the age of the father was four times that of his son, 5 years previous to the time when their ages were compared.

The question, taken in its general, or algebraic sense, demands the time, at which the age of the father was four times that of the son. In stating it, we supposed that the age of the father was to be augmented; and so it was, by the first supposition. But the conditions imposed by the second supposition, required the age of the father to be diminished, and the algebraic result conformed to this condition, by appearing with a negative sign. If we wished the result, under the second supposition, to have a positive sign, we might alter the enunciation by demanding, how many years since the age of the father was four times that of the

Reasoning from analogy, we establish the following general principles,

1st. Every negative value found for the unknown quantity in a problem of the first degree, will, when taken with its proper sign, verify the equation from which it was derived.

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

sense.

3d. The negative result shows that the enunciation is impossible, regarded in its arithmetical sense. The language of Algebra detects the error of the arithmetical enunciation, and indicates the general relation of the quantities.

4th. The negative result, considered without reference to its sign, may be regarded as 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 reciprocally.

106. As a further illustration of the change which an algebraic sign may produce in the enunciation of a problem, let us resume that of the laborer (page 76).

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

Now, it is plain that we can obtain immediately the values of x and y, in the last equations, by merely changing the sign of c in each of the values found from the equations above; this gives

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

The double sign ±, read plus or minus, and , is read, mi nus 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

107. When a problem has been resolved generally, that is, by means of letters and signs, it is often required to determine what the values of the unknown quantities become, when particular suppositions are made upon the quantities which are given. The determination of these values, and the interpretation of the peculiar results obtained, form what is called the discussion of the problem.

The discussion of the following question presents nearly all the circumstances which are met with in problems of the first degree.

108. Two couriers are travelling along the same right line and in the same direction from R' toward R. The number of miles travelled by one of them per hour is expressed by m, and the number of miles travelled 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 will be together.

At 12 o'clock suppose the forward courier to be at B, the other at A, and R to be the point at which they will be together. Then, AB = a, their distance apart at 12 o'clock.

Let

t =

and

X=

the number of hours which must clapse, before they come together;

the distance BR, which is to be passed over by the forward courier.

Then, since the rate per hour, multiplied by the number of

hours, will give the

Now, so long as m>n, t will be positive, and the problem will be solved in the arithmetical sense of the enunciation. 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.

But; if we suppose m< n, then, m n will be negative, and the value of t will be negative. How is this result to be interpreted?

It is easily explained from the nature of the question, which considered in its most general sense, demands the time when the couriers are together.

Now, under the second supposition, the courier which is in advance, travels the fastest, and 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 example, therefore, conforms to the general principle, that, if the conditions of a problem are such as to render the unknown quantity essentially negative, it will appear in the result with the minus sign, whenever it has been regarded as positive in the enunciation.

If we wish to find the distances AR and BR, passed over by the two couriers before coming together, we may take the equation

t=

a

m-n