size, the six will be changed to four. Required the circumference of each wheel.

30.

There is a number consisting of two digits; the number is equal to three times the sum of its digits, and if it be multiplied by three, the result will be equal to the square of the sum of its digits. Find the number.

31. A certain number of two digits contains the sum of its digits four times and their product twice. What is the number?

32. A person proposes to travel from A to B, either direct by coach, or by rail to C, and thence by another train to B. The trains travel three times as fast as the coach, and should there be no delay, the person starting at the same hour could get to B 20 minutes earlier by coach than by train. But should the train be late at C, he would have to wait there for a train as long as it would take to travel from C to B, and his journey would in that case take twice as long as by coach. Should the coach however be delayed an hour on the way, and the train be in time at C, he would get by rail to B and half way back to C, while he would be going by coach to B. The length of the whole circuit ABCA is 763 miles. Required the rate at which the coach travels.

XIV. DISCUSSION OF SOME PROBLEMS WHICH LEAD TO SIMPLE EQUATIONS.

186. We propose now to solve some problems which lead to Simple Equations, and to examine certain peculiarities which present themselves in the solutions. We begin with the following problem: What number must be added to a number a in order that the sum may be b? Let x denote this number; then,

This formula gives the value of x corresponding to any assigned values of a and b. Thus, for example, if a = 12 and b=25, we have x=25-12-13. But suppose that a = = 30 and b=24; then x=24-30-6, and we naturally ask what is the meaning of this negative result? If we recur to the enunciation of the problem we see that it now reads thus: What number must be added to 30 in order that the sum may be 24? It is thus obvious, that if the word added and the word sum are to retain their arithmetical meanings, the proposed problem is impossible. But we see at the same time that the following problem can be solved: What number must be taken from 30 in order that the difference may be 24 and 6 is the answer to this question. And the second enunciation differs from the first in these respects; the words added to are replaced by taken from, and the word sum by difference.

187. Thus we may say that, in this example, the negative result indicates that the problem in a strictly Arithmetical sense is impossible; but that a new problem can be formed by appropriate changes in the original enunciation to which the absolute value of the negative result will be the correct answer.

188. This indicates the convenience of using the word add in Algebra in a more extensive sense than it has in Arithmetic. Let x denote a quantity which is to be added algebraically to a; then the Algebraical sum is a +x, whether x itself be positive or negative. Thus the equation a+x=b will be possible algebraically whether a be greater or less than b. We proceed to another problem.

189. A's age is a years, and B's age is b years; when will A be twice as old as B Supposed the required epoch to be x years then by the question,

from the present time;

10. But

=

Thus, for example, if a 40 and b=15, then x=10. suppose a = 35 and b = 20, then x=-5; here, as in the pre

ceding problem, we are led to inquire into the meaning of the negative result. Now with the assigned values of a and b the equation which we have to solve becomes

35+x=40+ 2x,

and it is obvious that if a strictly arithmetical meaning is to he given to the symbols x and +, this equation is impossible, for 40 is greater than 35, and 2x is greater than x, so that the two members cannot be equal. But let us change the enunciation to the following: A's age is 35 years, and B's age is 20 years, when was A twice as old as B? Let the required epoch be x years from the present time, then by the question,

thus,

35-x=2 (20 - x) = 40 – 2x;

x = 5.

Here again we may say the negative result indicates that the problem in a strictly Arithmetical sense is impossible, but that a new problem can be formed by appropriate changes in the original enunciation, to which the absolute value of the negative result will be the correct answer.

We may observe that the equation corresponding to the new enunciation may be obtained from the original equation by changing x into - x.

190. Suppose that the problem had been originally enunciated thus: A's age is a years, and B's age is b years; find the epoch at which A's age is twice that of B. These words do not intimate whether the required epoch is before or after the present date. If we suppose it after we obtain, as in Art. 189, for the required number of years x = a -26. If we suppose the required epoch to be x years before the present date we obtain x=2b-a. If 2b is less than a, the first supposition is correct, and leads to an arithmetical value for x; the second supposition is incorrect, and leads to a negative value for x. If 2b is greater than a, the second supposition is correct, and leads to an arithmetical value for ; the first supposition is incorrect and leads to a negative value for x. Here we may say then that a negative result indicates that we made the wrong choice out of two possible supposi

tions which the problem allowed. But it is important to notice, that when we discover that we have made the wrong choice, it is not necessary to go through the whole investigation again, for we can make use of the result obtained on the wrong supposition. We have only to take the absolute value of the negative result and place the epoch before the present date if we had supposed it after, and after the present date if we had supposed it before.

191. One other case may be noticed. Suppose the enunciation to be like that in the latter part of Art. 189; A's age is a years, and B's age is b years, when was A twice as old as B? Let x denote the required number of years; then

hence,

a − x = = 2(b − x),
x=2b-a.

Now let us verify this solution. Put this value for x; then a-x becomes a− (2b—a), that is, 2a-26; and 2(b-x) becomes 2(b −2b+a), that is, 2a-2b. If b is less than a, these results are positive, and there is no Arithmetical difficulty. But if b is greater than a, although the two members are algebraically equal, yet since they are both negative quantities, we cannot say that we have arithmetically verified the solution. And when we recur to the problem we see that it is impossible if a is less than b; because if at a given date A's age is less than B's, then A's age never was twice B's and never will be. Or without proceeding to verify the result, we may observe that if b is greater than a, then x is also greater than a, which is inadmissible. Thus it appears

that a problem may be really absurd, and yet the result may not immediately present any difficulty, though when we proceed to examine or verify this result we may discover an intimation of the absurdity.

192. The equation a + x = 2 (b+x) may be considered as the symbolical expression of the following verbal enunciation. Suppose a and b to be two quantities, what quantity must be added to each so that the first sum may be twice the second? Here the words quantity, sum, and added may all be understood in Alge

braical senses, so that x, a, and b may be positive or negative. This Algebraical statement includes among its admissible senses the Arithmetical question about the ages of A and B. It appears then that when we translate a problem into an equation, the same equation may be the symbolical expression of a more comprehensive problem than that from which it was obtained.

We will now examine another problem.

193. A and B travel in the same direction at the rate of a and b miles respectively per hour. A arrives at a certain place P at a certain time, and at the end of n hours from that time B arrives at a certain place Q. Find when A and B meet.

Let c denote the distance PQ; suppose A and B to travel in the direction from P towards Q, and to meet at R at the end of x hours from the time when A was at P; then since A travels at the rate of a miles per hour, the distance PR is ax miles. Also B goes over the distance QR in x-n hours, so that QR is b(x−n) miles. And PR is equal to the sum of PQ and QR; thus,

We shall now examine this result on different suppositions as to the values of the given quantities.

I. Suppose a greater than b, and c greater than bn; then the value of x is positive, and the travellers will meet, as we have supposed, after A arrives at P. For when A is at P, the space which B has to travel before he reaches Q is bn miles, and since bn is less than c, it follows that when A is at P he is behind B; and A travels more rapidly than B, since a is greater than b. Hence A must at the end of some time overtake B,