A and C are worth is equal to q; 3d, that what C is worth added to n times what A and B are worth is equal to r. This un This question can be resolved in a very simple manner, by intro ducing an auxiliary unknown quantity into the calculus. known quantity is equal to what A, B and C are worth. 28. Find the values of the estates of six persons, A, B, C, D, E, F, from the following conditions: 1st. The sum of the estates of A and B is equal to a; that of C and D is equal to b; and that of E and F is equal to c. 2d. The estate of A is worth m times that of C; the estate of D is worth n times that of E, and the estate of F is worth p times that of B. This problem may be resolved by means of a single equation, involving but one unknown quantity. Theory of Negative Quantities. Explanation of the terms, Nothing and Infinity. 104. The algebraic signs are an abbreviated language. They point out in the shortest and clearest manner the operations to be performed on the quantities with which they are connected. Having once fixed the particular operation indicated by a particular sign, it is obvious that that operation should always be performed on every quantity before which the sign is placed. Indeed, the principles of algebra are all established upon the supposition, that each particular sign which is employed always means the same thing; and that whatever it requires is strictly performed. Thus, if the sign of a quantity is +, we understand that the quantity is to be added; if it is we understand that it is to be subtracted. For example, if we have -4, we understand that this 4 is to be subtracted from some other number, or that it is the result of a subtraction in which the number to be subtracted was the greatest. If it were required to subtract 20 from 16, the subtraction could not be made by the rules of arithmetic, since 16 does not contain 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 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 essen. tial 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 a+b=a, whence a-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. 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. Let us now suppose a=45, and b=15: then x= 45-60 -- -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 equations. But if we suppose that the labourer, instead of receiving, owes a sum c, the equations will then be 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 c by d; The equations then become ( x + y = n Iax- by=d and they only differ from those of the first enunciation by having d in the place of c. We would, therefore, necessarily find And by substituting -c for d, we have bn+(-c) an-(—c) X a+b ; y= a+b ; The results, which agree to both enunciations, may be comprehended in the same formula, by writing |