This result, 0=108, being manifestly absurd, it follows that no value of can verify equation 1, and, by consequence, that none can satisfy the question.

For this reason the denomination of absurd or impossible is applied to the equation and the question.

The impossibility may be rendered apparent in equation 1 by reducing the like terms of its first member;

Consequently the two members must always differ by 9, whatever may be the value assigned to x.

This impossibility is very different from that of Article 73.

The question of that Article was impossible, not because the value of ± failed to verify the equation, but because it was negative, and the question excluded a negative value; but in the present question the impossibility is absolute, for no quantity of any kind can give the equality which is established by its enunciation.

78. Problem 2. It is required to find a number, x, such that (x+10) 2(x+20), 5(x-34) + may be equal to 2(x−5). 6 The equation of the problem is,

+

3

+

+

2

3

=2(x-5)

1

x+10, 2(x+20), 5(x−34).
6
..3x+30+4x+80+5x-170=12x-60.

Transposing, 3x+4x+5x-12x=170-30-80-60, or 0=0.

To explain this result, let the first member of equation 1 be reduced to the most simple form; it is

3x+30+4x+80+5x-170
6

12x-60
6

́, or 2x—10, or 2(x−5).

Equation 1 is thus reduced to 2(x−5)=2(x−5), an expression in which if a is replaced by any number whatever the equation is verified.

Whence equation I is an equality which subsists, whatever be the value of x.

In all such cases the equation and the question from which it is deduced are called Indeterminate.

79. Problem 3. It is required to find two numbers such that one half of the first may be equal to one third of the second plus 5, and the second equal to three times one half of the first less 5 ?

x y
2

Let a represent the first number, y the second, then +5 (1) and

Substituting in the 1st equation the value of y, which is given by the 2d,

1/3x

32 +5,

x x 5

2—2—3+5;

...3x=3x-10+30,

3x-3x=30-10,

and ...0=20.

From this absurd result, 0=20, it is evident that no values of x, y can satisfy the two equations; consequently the problem must involve impossible conditions.

Considering equations 1 and 2 individually;

The first, by clearing fractions and transposing terms, becomes

The question therefore requires that the same quantity, 3x-2y, shall be equal, first, to 30, and second, to 10, which is impossible.

Each of the equations, 3x-2y=30, 3x-2y=10, taken singly, is possible, and susceptible of an indefinite number of solutions; but the two conditions of the problem cannot coexist.

Conditions such as these, and also the equations by which they are expressed, are said to be Contradictory or Incompatible.

Let the problem under consideration be changed to this: find two numbers such that one half of the first shall be equal to one third of the second plus 5, and the second equal to three times the half of the first less 15. The equations of this problem are,

Substituting in equation 1 the value of y, given by equation 2,

Consequently the two equations are indeterminate (Art. 78).

This conclusion may be deduced from equations 1 and 2, by removing the denominators, and transposing both the unknown quantities to the first members of the equation.

x

For since =3+5,

3x=2y+30,

and 3x-2y=30.

3x

Since y= -15,

2
2y=3x-30,
and 3x-2y=30.

Whence equations 1, 2, being reducible to the single equation 3x-2y=30, it follows that the two conditions indicated in the enunciation differ in appearance only.

The two equations, therefore, as well as the problem from which they are deduced, are indeterminate.

80. The preceding problems illustrate the cases of exception mentioned in Article 57.

In all cases of exception the results express either impossibility or indetermination.

To establish this proposition: it is evident that, from the processes of elimination employed, there can happen only one of the three contingencies: 1st. From any number of equations to the same number of unknown quantities, a single equation of the first degree containing only one unknown quantity is deduced; from this equation a single value of the remaining unknown quantity is obtained; then, by substituting this value in the equation which contains this and another unknown quantity, a single value of the second unknown quantity is obtained; and, in the same manner, by returning through the equations containing 3, 4, ... unknown quantities,

a single value of each unknown quantity is determined. This is the case of the Rule in Article 55.

Or, 2d. In the course of the operation or at its conclusion an equation is found in which the unknown terms destroy each other, and the remaining terms of the two members are given quantities, different from each other.

There exists in this case an evident absurdity, and the conclusion is, that whatever values are given to the unknown quantities, neither the primitive equations nor the question from which these equations are deduced can be satisfied. The problem of Article 77 falls under this case.

Or, 3d. If neither the first nor the second case occurs, one or more equations must be found in which the unknown quantities destroy one another, and the two members are equal to the same known quantity.

In this case any values may be given to the unknown quantities contained in the last equation, and then by means of these the values of the other unknown quantities may be found. The problem of Article 78 falls under this case.

In this case there exists a true indetermination in the primitive equations, and by consequence in the problem of which they express the conditions.

If in considering two equations to two unknown quantities an absurd equality is deduced, since it is evident that each equation taken individually is possible, it ought to be concluded that the two equations are contradictory or incompatible.

If the result is an identical equation, the conclusion is that one of the equations is always verified when the other is verified, or that the one equation is a consequence of the other.

81. The number of equations and unknown quantities being three, there will be impossibility if one equation is incompatible with either of the others, or with the equation deduced from them, without being so with either equation considered separately;

And there will be indetermination if one equation is a consequence of either of the others, or if two of the equations are consequences of the third, or if one equation is a consequence of the equation deduced from the other two without being a consequence of either of them taken separately.

When the number of equations is greater (but always equal to the number of unknown quantities) impossibility and indetermination arise from more varied causes; but in general, if it is required to find whether an equation is incompatible with or a consequence of several others, the values of the unknown quantities in the latter equations are to be found in terms of one unknown quantity and of the known quantities of the equations, and these values are to be substituted for the unknown quantities in the first equation (with the exception of the unknown quantity not eliminated from the other equations); then, if the result after reduction is an absurd equality, the first equation is incompatible with the system of the other equations; and if the result is an identity, the first is a consequence of the system of the other equations.

82. Besides the cases of impossibility presented by equations, there are others which may arise from the conditions of a question. For example, it may be required by the enunciation of a problem that the value of an unknown quantity shall be a whole number, and the value obtained from the equations of the problem may be a fraction; or it may be required that the value of the unknown quantity shall be less than a given number, and the result obtained may be greater than that number. . . . . . In general, the enunciation of a problem may impose upon one or upon several unknown quantities certain conditions which it is impossible to express by equations; and in these cases, after obtaining the values of the unknown quantities, it remains to try whether they satisfy all these conditions.

EXAMPLE OF THE DISCUSSION OF A PROBLEM OF THE FIRST

DEGREE.

83. If an equation of the first degree, in which the given quantities are represented by letters, is resolved, the value of the unknown quantity (assuming, for the sake of simplicity, that there is only one) is expressed by a formula which indicates the operations to be performed upon the given quantities.

Now, the given quantities (or the letters by which they are represented) may be supposed to receive all possible values, and it may be required to determine what, under the different hypotheses, are the values of the unknown quantity, x.

The determination of the different values of x, and the interpretation of the remarkable results which are obtained, constitute what is termed the Discussion of a problem.

To illustrate the manner of conducting the discussion of a problem, let two bodies, B, B', move uniformly in the indefinite straight line XY with given velocities, v, v', and in the direction XY; also, let B reach the point A h hours before B' reaches A'; and let the distance AA' be d. It is required to find in what point of the line XY the bodies B, B' are in conjunction.

B

B'

Suppose that the point P between A' and Y is the place in which B, B' come together or are in conjunction: express A'P by x, and therefore AP by d+x.

The times taken by B, B' respectively to describe the distances d+x and d+x r with the given velocities, v, v', are and;

v

And since B reaches A h hours before B' reaches A', the time

v

In this solution it is assumed that B, B' come together between A' and Y; they may, however, come together either before or after reaching A'. Both cases are satisfied by considering x to be either additive or subtractive (Art. 76. 3); and, interpreted thus, equation 2 is perfectly general.

Discussion:

1st. The numerator and denominator of the value of x are both positive, when at the same time v>v', d>vh.

Whence the value of x is positive, and, by consequence, B, B' come together between A' and Y, as has been assumed.

This conclusion is in conformity with that which is obtained from the question itself; for vh is the distance described by B during the time, h, which elapses between the arrival of B at A and that of B' at A'.

Now, to suppose v>v' and d>vh is to admit that B moves faster than B', and that B' has reached A' before B reaches A'; consequently B must overtake B' on the Y side of A'.

2d. When at the same time v<v' and d<vh, the value of x is positive; and consequently B, B' still come together on the Y side of A'.

In effect it is evident in this case that B moves slower than B', and that it has passed the point A' before B' has reached A'; therefore B must be overtaken by B' on the Y side of A'.

3d. When at the same time v<v′ and d>vh,

or v>v' and d<vh, the value of x is negative.

This negative value of x indicates that B, B' come together on the X side of A', and not on the Y side (as was assumed), and at a distance from A' marked by the absolute value of r; that is, by the value of r considered without regard to its sign.

Returning, as before, to the question when d>vh, B' reaches A' before B reaches A'; and when v'>v, B' moves faster than B. Consequently, since B' reaches A' before B, and also moves faster than B, the bodies B, B' cannot be in conjunction on the Y side of A'.

Again, when d<vh, B has passed A' before B' has reached A'; and when v>', B moves faster than B'. Therefore, in this hypothesis also, the conjunction cannot take place on the Y side of A'.

Consequently, under both the hypotheses of case 3, the conjunction of B, B' must take place on the X side of A'.

It has been supposed that B, B' are in conjunction on the Y side of A ́. Now the negative value of x proves that in the 3d case this supposition is inadmissible, but does not expressly point out whether the conjunction takes place on the X side of A, or between A and A'.

If the conjunction takes place in P' on the X side of A, make A ́P′=x ; then the time taken by B to move from P' to A is and by B' to move

x-d

v

from P' to A' is The time elapsed between the arrival of B at the point A, and that of B' at the point A', is therefore expressed by

x-d

v

and as the question requires that this time shall be equal to h,

x x-d
=h

v

v

3

If the conjunction takes place between A' and A in the point P", and the unknown quantity A' P" is denoted by x, the time taken by B to move d-x from A to P" is and the time taken by B' to move from P" to A' is

Therefore the time elapsed between the arrival of B in A and that of B' d-xx'

; and by the question,

in A' is

v

which is the left member of equation 4, it follows that equations 3, 4 are the same. Therefore equation 4 ought to determine the point of conjunction, wherever may be its position on the X side of A'. And since equation 1 becomes equation 4 by changing x intox, it is to be concluded that every positive value which satisfies equation 4 will be given, but with the sign by equation 1.

Whence equation 1 is of itself sufficient to determine in all cases the point of conjunction of the bodies B, B', care only being taken to carry the negative values of x to the left of A'.