20; nor indeed can it be entirely performed by Algebra. We write the numbers for subtraction thus,

16-20-16-16-4--4.

By decomposing -20 into -16 and -4, the -16 will cancel the +16, and leave -4 for a remainder.

We thus indicate that the quantity to be subtracted exceeds the quantity from which it is to be taken, by 4.

To show the necessity of giving to this remainder its proper sign, let us suppose that the difference of 16-20 is to be added to 10. The numbers would then be written

105. If the sum of the negative quantities in the first member of the equation, exceeds the sum of the positive quantities, the second member of the equation will be negative, and the verification of the equation will show it to be so.

and we make a=15 and b=18, c will be =-3. Now the essential sign of c is different from its algebraic sign in the equation. This arises from the circumstance, that the equation a-b=c expresses generally, the difference between a and b, without indicating which of them is the greater. When, therefore, we attribute particular values to a and b, the sign of c, as well as its value, becomes known.

We will illustrate these remarks by a few examples.

1. To find a number which, added to the number b, gives for a sum the number a.

Let x= the required number.

Then, by the condition x+b=a, whence x=a-b.

This expression, or formula, will give the algebraic value of x in

all the particular cases of this problem.

For example, let a=47, b=29, then x=47-29=18.

Again, let a=24, b=31; then will x=24-31=-7.

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

Considered arithmetically, the problem with these values of a and b, is impossible, since the number b is already greater than 24. Considered algebraically, however, 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.

x=

Let the required number of years. Then a+x= the age of the father and b+x= the age of the son ed time.

Hence, by the question a+x=b+x;

4

}at

at the end of the requir

-=b+x; whence x=

Suppose a=54, and b=9: then x=

54-36 18

3

3

6.

a-46
3

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

45-60.
3

Let us now suppose a=45, and b=15: then x= -=-5.

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

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

Now, the positive result which has been 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 most 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 his son. If x= the number of years, we shall have

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. If the enunciation is to be understood in its arithmetical sense, in which the quantities referred to are always supposed to be positive, then this value, considered without reference to its sign, may be considered as the answer to a problem, of which the enunciation only dif. fers from that of the proposed problem in this, that certain quantities which were additive. have become subtractive, and reciprocally.

106. Take for example the problem of the labourer (Page. 88). Supposing that the labourer receives a sum c, we have the

But if we suppose that the labourer, instead of receiving, owes a

By changing the signs of the second equation.

Now it is visible that we can obtain immediately the values of x

and y, which correspond to the preceding values, by merely chang

ing the sign of c in each of those values; this gives

To prove this rigorously, let us denote -c by d;

{ x+y=n
(ax-by=d

and they only differ

The equations then become from those of the first enunciation by having d in the place of c. We would, therefore, necessarily find bn+d an-d a+b , y= a+b And by substituting -c for d, we have bn+(-c) a+b

X

; y=

an-(-c)
a+b

an+c a+b; y= a+b

;

The results, which agree to both enunciations, may be compre

hended in the same formula, by writing

The double sign ± is read plus or minus, the superior signs correspond to the case in which the labourer received, and the inferior signs to the case in which he owed a sum c. These formulas comprehend the case in which, in a settlement between the labourer and his employer, their accounts balance. This supposes c=0, which gives

107. When a problem has been resolved generally, that is, by representing the given quantities by letters, it may be required to determine what the values of the unknown quantities become, when particular suppositions are made upon the given quantities. 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' towards 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.

the number of hours which must elapse, 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 distance passed over by each, we have,