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CHAPTER X.

DISCUSSION OF PROBLEMS INVOLVING SIMPLE EQUATIONS.

INEQUALITIES. 159. To discuss a problem or an equation is to determine the values which the unknown quantities assume for particular hy. potheses made upon the values of the given quantities, and to interpret the peculiar results obtained. We have seen that if the sum of two numbers is represented by a, and their difference by b, the greater number will be expressed by

2
a-b
Here a and 5

may

have 2

values whatever,

any and still these formulæ will always hold true. It frequently happens that, by attributing different values to the letters which represent known quantities, the values of the unknown quantities assume peculiar forms, which deserve consideration.

a+b

the less by a

160. We may obtain five species of values for the unknown quantity in a problem of the first degree:

1st. Positive values. 2d. Negative values.

0 3d. Values of the form of zero, or

A 4th. Values of the form of

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5th. Values of the form of

0 We will consider these five cases in succession.

161. 1st. Positive values are generally answers to problems in the sense in which they are proposed. Nevertheless, all positive values will not always satisfy the enunciation of a problem. For example, a problem may require an answer in whole numbers, in which case a fractional value of the unknown quan. tity is inadmissible. Thus, in Prob. 17, page 93, it is implied that the value of x must be a whole number, although this condition is not expressed in the equations. We might change the data of the problem, so as to obtain a fractional value of x, which would indicate an impossibility in the problem proposed. Problem 43, page 97, is of the same kind; also Prob. 7, page 109.

If the value obtained for the unknown quantity, even when positive, does not satisfy all the conditions of the problem, the problem is impossible in the form proposed.

162. 2d. Negative values.

Let it be proposed to find a number which, added to the number 6, gives for a sum the number a. Let x denote the required number; then, by the conditions of the problem,

6+x=a; whence

x=a-6. This formula will give the value of x corresponding to any assigned values of a and b.

For example, if a=7 and b=4, then

x=7—4=3, a result which satisfies the conditions.

But suppose that a=5 and b=8, then

x=5-8=-3. We thus obtain for x a negative value. How is it to be interpreted ?

By referring to the problem, we see that it now reads thus: What number must be added to 8 in order that the sum may be 5? It is obvious that if the word added and the word sum are to retain their arithmetical meanings, the proposed problem is impossible. Nevertheless, if in the equation 8 + x=5 we substitute for +ä its value - 3, it becomes

8-3=5, an identical equation; that is, 8 diminished by 3 is equal to 5, or 5 may be regarded as the algebraic sum of 8 and - 3.

The negative result, x=-3, indicates that the problem, in a strictly arithmetical sense, is impossible; but, taking this value of with a contrary sign, we see that it satisfies the enunciation when modified as follows: What number must be subtracted from 8 in order that the difference may be 5 ? The second enunciation differs from the first. only in this, that we put subtract for add, and difference for sum. If we wish to solve this new equation directly, we shall have

8-x=5; whence

x=8-5, or 3.

163. For another example, take Problem 50, page 98. The age of the father being represented by a, and that of the son

a-nb by b, then

will represent the number of years before the

n-1 age of the father will be n times that of the son.

a=54, b=9, and n=4; then

54-36 18

=*=6. 3 3

Thus, suppose

But suppose

X=

This value of x satisfies the conditions understood arithmetically; for if the father was 54 years old, and the son 9 years, then in years more the age of the father will be 60 and the son 15; and we see that 60 is 4 times 15.

a=45, b=15, and n=4;

45-60 - 15 then

-5. 3

3 Here again we obtain a negative result. How are we to interpret it?

By referring to the problem, we see that the age of the son is already more than one fourth that of the father, so that the time required is already past by five years. The problem, if. taken in a strictly arithmetical meaning, is impossible. But let us modify the enunciation as follows:

The age of the father is 45 years; the son's age is 15 years; how many years since the age of the father was four times that of his son ?

The equation corresponding to this new enunciation is

15-x=45—x

4 whence

60—4x=45-X; and x=5, a result which satisfies the modified problem taken in its arithmetical sense.

From this discussion we derive the following general principles :

1st. A negative result found for the unknown quantity in a prob lem of the first degree indicates that the problem is impossible, if understood in its strict arithmetical sense.

2d. This negative value, taken with a contrary sign, may be regarded as the answer to a problem whose enunciation only differs from that of the proposed problem in this, that certain quantities which were ADDED should have been SUBTRACTED, and vice versa.

164. In what case would the value of the unknown quantity in Prob. 20, page 94, be negative?

Ans. When nm. Thus, let m=20, n=25, and a=60 miles;

60 60 then

20-25 -5

=-12. To interpret this result, observe that it is impossible that the second train, which moves the slowest, should overtake the first. At the time of starting, the distance between them was 60 miles, and each subsequent hour the distance increases. If, however, we suppose the two trains to have been moving uniformly along an endless road, it is obvious that at some former time they must have been together.

This negative result indicates that the problem is impossible if understood in its strict arithmetical sense. But if the problem bad been stated thus:

Two trains of cars, 60 miles apart, are moving in the samo direction, the forward one 25 miles per hour, the other 20. How long since they were together? The problem would have furnished the equation

25x=20x+60; whence

X= +12.

If we wish to include both of these cases in the same enun. ciation, the question should be, Required the time of their being together, leaving it uncertain whether the time was past or future.

EXAMPLES.

part by 16?

1. What number is that whose fourth part exceeds its third

Ans. -192. How should the enunciation be modified in order that the result may be positive?

2. The sum of two numbers is 2, and their difference 8. What are those numbers ?

Ans. -3 and +5. How should the enunciation be modified in order that both results may be positive?

3. What fraction is that from the numerator of which if 4 be subtracted the value is one half, but if 7 be subtracted from the denominator its value is one fifth ?

-5 Ans.

-18 How should the enunciation be modified in order that the problem may be possible in its arithmetical sense ?

4. Find two numbers whose difference is 6, such that four times the less may exceed five times the greater by 12.

Ans. -42 and -36. Change the enunciation of the problem so that these numbers, taken with the contrary sign, may be the answers to the modified problem.

165. 3d. We may obtain for the unknown quantity values of

0 the form of zero, or

A In what case would the value of the unknown quantity in Prob. 20, page 94, become zero, and what would this value signify?

Ans. This value becomes zero when a=0, which signifies that the two trains are together at the outset.

In what case would the value of the unknown quantity in Prob. 50, page 98, become zero, and what would this value signify?

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