qrs. of

34th. Given the value of 24 qrs. of wheat, 9 qrs. of rye, and 15 barley 36 £. 14 sh.; the value of 9 qrs. of wheat, 30 qrs. of rye, and 21 qrs. of barley=40£. 12 sh.; and the value of 18 qrs. of wheat, 27 qrs. of rye, and 39 qrs. of barley=56 £. 10 sh.; to find the price of a quarter of each kind of grain ?......Ans. Wheat 18 sh. 8 d., barley 14 sh., and rye 10 sh. 8 d. per qr.

35th. A cistern is filled by 3 pipes, A, B, C; the pipes A, B together fill the cistern in 70 minutes; A, C, together, in 84 minutes; B, C, together, in 140 minutes. In what time will each pipe fill the cistern, and in what time will it be filled if all three pipes are open?

Ans. By A in 105 min., by B in 210 min., by C in 420 min., and by A, B, C together in 60 min.

36th. Supposing that 32 lbs. of sea water contain 1 lb. of salt, how much fresh water must be mixed with these 32 lbs. in order that 32 lbs. of the mixture may contain only 2 oz. of salt, or of the former quantity?.....Ans. 224 lbs.

37th. A number is expressed by 3 figures; the sum of the figures is 13; the figure which expresses the simple units of the number is triple the figure which expresses the hundreds; and if 396 is added to the number the sum is the number reversed. Required the number ?....................................................................Ans. 256. 38th. Three brothers, A, B, C, buy a house for 2000£.; C can pay the whole price if B give him half of his money; B can pay the whole price if A give him one third of his money; A can pay the whole price if C give him one fourth of his money. How much has each ?

Ans. A 1680 £., B 1440 £., C 1280 £.

39th. A silversmith has three bars composed of silver, copper, and tin mixed in different proportions. The pound (avoirdupois) of the first bar contains 7 oz. of silver, 3 oz. of copper, and 6 oz. of tin; the pound of the second contains 12 oz. of silver, 3 oz. of copper, and 1 oz. of tin; and the pound of the third contains 4 oz. of silver, 7 oz. of copper, and 5 oz. of tin. How much of each of these 3 bars must be taken to form a fourth, the pound weight of which shall contain 8 oz. of silver, 33 oz. of copper, and 44 oz. of tin?

Ans. 8 oz. of the first bar, 5 oz. of the second, and 3 oz. of the third. 40th. A father leaves his property to be divided among his three children in the following manner: the first is to receive the sum a, plus the nth part of the remainder; the second is to receive the sum 2a, plus the nth part of what remains after the subtraction of the first part augmented by 2a; the third is to receive the sum 3a, plus the nth part of what remains after the subtraction from the property of the first part plus the second part plus 3a; the property is thus entirely divided. What is its value?

Ans.

a(6n2-4n+1) n2-2n+1

OF THE INTERPRETATION OF NEGATIVE RESULTS IN EQUATIONS AND PROBLEMS OF THE FIRST DEGREE.

73. The employment of algebraic symbols in the resolution of problems leads, in some instances, to remarkable results.

These are most intelligibly illustrated and explained through the medium of examples.

Let it be supposed, therefore, that a workman was employed 13 days in summer at certain wages, and 17 days in winter at 2 sh. less per day than in summer; that in summer he forfeited 22 sh. for waste of materials, and in winter received 28 sh. for extra work, and that he each time received the same sum; and let it be required from these conditions to find the workman's day's wages in summer?

To resolve the question,

Let x sh. the day's wages in summer;

..x-2 sh. the day's wages in winter :

Also 13x sh. wages of 13 days at x sh. per day;

13x-22 sh. sum received in summer;

17(x-2)=17x-34 sh.=wages of 17 days at x-2 sh. per day,
and 17x-34+28 or 17x-6=sum received in winter.

But, by the question, these sums are equal:

17(x−2)+28=13x—22, or
17x-6=13x-22

Therefore {17(x

or}

1

Transposing, 17x-13x=6-22,

or 4x=-16,

16

=-4. 4

The result which is obtained, namely, that the workman's day's wages in summer is —4 sh., has no intelligible meaning.

It is to be inferred from this result, which is termed a Negative solution, that the enunciation of the problem contains impossible conditions.

It is evident, from Article 47, that the changes which are made to obtain the equation x=-4 from the equation

17(x-2)+28=13x-22,

cannot alter the value of the unknown quantity. Now the equation x=-4 is verified by substituting -4 for r (since-4=-4), and cannot be verified by any other number; therefore the equation 17(x-2)+28=13x−22 can also be verified by this value alone, of x.

Now, if there exists a value of x which answers the question, this value ought of necessity to verify equation 1; and since the verification cannot be made but by a negative value, it is to be concluded that the question is impossible.

But since the negative value x=-4 verifies equation 1, which is the immediate expression of the conditions of the problem; that is,

since 17(x-2)+28=13(x)−22

or 17(-4-2)+28=13(-4)—22

if x is replaced in equation 1 by -x, it becomes

17(−x−2)+28=13(-x)-22 .

3

and since x-4, and consequently -x=4, it is evident that the substitution of -4 for x in equation 1, and the substitution of 4 for -x in equation 3, must give precisely the same result.

Therefore, since equation 1 ought to be verified by the value x=-4, equation 3 must be verified by the value x=4. Hence every question whose enunciation is expressed by equation 3 will have for solution x=4.

But as it is absurd to suppose a question which leads to an equation containing only negative quantities in its second member, it becomes necessary to change all the signs of equation 3, and to write it thus,

In order, therefore, that the question proposed may have the solution x=4, it is sufficient to modify its enunciation in such a manner that the workman's day's wages in winter may be 2 sh. greater than in summer; that the sum gained in the 13 days of summer may be augmented by 22 sh., instead of being diminished by that sum; and that the sum gained in the 17 days of winter may be diminished by 28 sh., instead of being augmented by that

sum.

With these changes, the question is

A workman was employed 13 days in summer at certain daily wages, and 17 days in winter at 2 sh. a day more than in summer. In summer he received a gratuity of 22 sh. for extra work, and in winter forfeited 28 sh. for waste of materials. Each time he received the same sum. What were his daily wages in summer?

It is evident that the question, thus changed, leads to equation 4, which admits the same value, x=4, as equation 3; consequently, this value resolves the question in the precise sense of the new enunciation.

It is to be especially noticed that the known quantities of the first enunciation have remained without alteration of absolute value in the new; only some of them which at first were additive have been rendered subtractive, and conversely.

In general, if the enunciation of a problem requires that the unknown quantity shall have a positive value, and the equation gives a negative value, x=—a, it is to be concluded that the problem is impossible.

To rectify the enunciation, x is replaced in the equation by -x (by which means the signs of certain terms are changed), and, without altering the data, the manner of regarding them is modified in such a manner that the changes of signs of the equation correspond exactly to this modification.

Then it is certain, without resolution of the new equation, that the positive value x=a, corresponds to the new enunciation, unless it involves some conditions which are not expressible by an equation; as, for example, that the unknown quantity shall be a whole number.

In conclusion, this rectification may be made in many different ways; but it is always necessary to model the new as exactly as possible after the primitive enunciation.

74. Two couriers, A, B, who travel uniformly, A at the rate of 10 miles, and B at the rate of 8 miles an hour, start at the same time for London from Newcastle and York; the distance of Newcastle from London being 270 miles, and of York from London 190 miles, it is required to find at what point in the road A overtakes B?

Suppose that A overtakes B at P, and make PL=x, Then, at P, A has travelled 270-x miles, and B 190-2 miles.

As A and B start at the same time, the number of hours denoted by 270-x

190-x

10

must be equal to the number denoted by

The answer is negative; and yet, since A, who is behind B at starting, travels two miles an hour faster than B, A must overtake B, if both proceed far enough.

In forming the equation, it is assumed that A overtakes B between York and London; and in this assumption exists the absurdity which is pointed out by the negative result.

For the purpose of avoiding this absurdity it is necessary to indicate as the place of meeting some point, P', beyond L; then making LP'=x, the equation to be resolved is

or

=

2

Equation 2 is equation 1, with - substituted for r; therefore, since equation 1 is satisfied by the value x=-70, equation 2 must be satisfied by the value x=70 (Art. 73). In this manner it is found that the place where A overtakes B is 70 miles beyond London.

Hence it follows that equation 1 gives the solution of the problem, provided the negative value of x is carried along the production of the line NYL; that is, on the side of L opposite to that on which it was first assumed to be situated.

The distance YP may be taken as ; then to obtain PL it becomes necessary to subtract YP from YL. If the remainder is negative, this arises from the circumstance that the place in which A overtakes B is a point, P', situated beyond L, at a distance, LP', equal to the remainder taken additively.

75. Two couriers start at the same time from two towns situated on the same road along which they travel. The distance between the two towns and the number of miles travelled in the hour by each courier are known, and hence it is required to find the place of their coming together?

Expressed thus, the question presents two distinct cases:

1st. That in which the couriers go in the same direction.

2d. That in which they go in opposite directions.

1st. Let one courier start from A, the other from B; let both proceed in the same direction from A to C, and suppose that P is the place where the courier from A overtakes the courier from B.

Also, let d denote the given distance AB;

m the number of miles travelled per hour by the courier from A,
m'

B, a the unknown distance, AP, gone by the courier from A before he overtakes the courier from B. Then the distance travelled by the courier from B is AP-AB, or x-d.

x

Dividing x by m, expresses the time taken by the courier from A to reach P;

2d. Let the couriers, starting at the same time, one from A, the other from B, advance to meet each other; and let the place of meeting be P.

Adopting the notation of the first case,

and observing that if AP=x, BP=d-x, the two equivalent expressions of

time are

Equation 2 differs from equation 1 in respect only that m' is replaced by

x-d

-m'; for in replace m' by —m' then,

Whence equation 1 is sufficient for the solution of both cases of the problem, provided it is agreed to regard the rate of travelling per hour of the courier from B as changing its sign when the direction in which the courier travels is changed.

From the remarks on case 2, it appears that with the help of negative quantities may be expressed by a single equation, and consequently in one formula, many cases of the same question, each of which would at first appear to require a distinct solution.

For this reason negative quantities are constantly employed in algebraic investigations.

76. It is not possible to include under any general description the class of questions in which the solution of one case may, by a change of signs, be made to include other cases; but all questions in which it is necessary to consider two or more cases which differ from each other, in respect that quantities which are additive for one or more cases are subtractive for another or others, are comprehended in it.

Such are questions in which the same letter is to be regarded as representing a quantity positive or negative, according as it expresses a gain or a loss, a motion in one direction or in a contrary direction, distance on the right of a fixed point or on the left, &c. &c.

The conclusions to be deduced from Articles 73, 74, and 75 are,

1st. The negative value of an unknown quantity may indicate an impossible condition in the enunciation of a question. If so, it is sufficient in this enunciation to give contrary meanings to certain of the quantities, to remove the impossibility; and the negative value before obtained, being taken positively, will give the solution of the amended question.

2d. The negative value of an unknown quantity may also indicate an erroneous supposition; and in this case it gives the solution of the question, provided the quantity which it represents is taken with a meaning contrary to that which had been at first given to it.

3d. If certain quantities are considered positive or negative, according as they are taken to have one meaning or another meaning directly contrary, the different cases of the same question may be combined in a single equation or in one formula.

Example. A father is 45 years of age, his son is 15; in how many years will the age of the son be one fourth of the age of the father? The answer -5 shows that five years ago the age of the son was one fourth of the age of the father.

In effect 15-5-10; 45-5=40.
10=4 of 40.

OF IMPOSSIBILITY AND INDETERMINATION IN EQUATIONS AND PROBLEMS OF THE FIRST DEGREE.

aug

77. Problem 1. It is required to find a number such that of it mented by 75, plus of it diminished by 35, may be equal to of it augmented by 49?

Let x denote the number; then, by the question,

3x
+49