(10) x+y+6z, (a+b)(x+y)+(c—d)z, a(x+y)—dz, z—(a—b)(x+y), 3(x+y)—4. 23. It must be observed, in reference to this tenth example, that the coefficient (a—b) in the last expression but one, being preceded by the minus sign, is subtractive; that is, the quantity to be subtracted is a all but b; so that, if the whole of a be subtracted, b must afterwards be added to correct the error. And generally, whenever quantities within brackets are to be subtracted, those of them only which have the plus sign are actually to be taken away and the others added; because the notation implies that the real amount to be taken away is the sum of all the positive quantities, all but the sum of all the negative quantities; so that, if all the positive quantities be first taken away, without any abatement, then all the negative quantities must afterwards be added. It is of such importance to be always mindful of this in the management of bracketed quantities, that it may be prudent to give an illustration or two of our meaning. Taking the example just referred to, the coefficients of x+y are 1, a+b, a, —(a—b), and 3; and these are the same as 1, a+b, a, -ab and 3; the sum of which is a+26+4; and this is the coefficient of (x+y) in the result of example 10. Again, -(a-b+c—d—e), is the same as —a+b―c+d+e: for the expression indicates that a all but b, and c all but d and e, are to be subtracted; so that, if the whole of a be subtracted, b must be then added to correct the error; and if the whole of c be subtracted, d and e must both be added for a similar purpose: thus a bracketed quantity, preceded by a minus sign, is equivalent to the same quantity, without brackets, when the signs of all the individual terms are changed. And this is a principle that must be carefully remembered. plus sign, prefixed to a bracketed quantity, leaves the signs of the component terms undisturbed, though the brackets be removed. From thus understanding the office and influence of the sign prefixed to a bracketed quantity, the learner will be able to take any number of the terms of a compound expression, inclose them within brackets, and prefix to the bracket either sign he chooses, without committing error: thus, he might write the first of the following forms in either of the ways here subjoined, without altering the value or meaning of the expression: A 3a-2bx-4cy+d-4-3a-(2 bx +4cy-d+4)=3a+ (d-2bx-4cy-4)=3a-2bx-(4cy-d+4)=3a-2 bx4 cy+(d-4)=3a-2bx-4cy-(4-d) as is sufficiently obvious from what has already been said. (11) (a+b—c) x+4y—6, (2a + 3 c) x —(d—3)y, -(4-2c)x+2y-1. (12) (a2 —b2) (x2 —y2)+3 xy, 5 xy — (b2 — a2) (x2 — y2), (4 a2—3 b2)(x2—y3)—xy. (13) (m+4)√x-(n−3)√y+√z, 3√z—(n−1)√y+ (2-3 m)√x, -(a—b)√z. (14) √a+b (x—y)3—2(x—y), (a+b)*(x—y)3—(a—2) (x—y), (3—2a)(x—y)+ỷ x—y. (15) a—(b2-c2+d—e), 3a+26—(c2—d—e), 5c+e— (d-5b-4a), 2b2+c2-4a-(3 e-6d); the sum to be expressed without brackets. Perhaps the easiest way for the learner to proceed in this example is to remove the brackets from each of the proposed expressions at first, and then to add together the like quantities; and the same plan is recommended in the following final example. (16) a—(b—2c—d)—{3a—(4b—c)—d}, 5{2a—(d— 3a+c)}, 2{a-3(b-c)}; the sum to be expressed free of brackets. In proceeding with this example, as recommended above, the learner will probably find it the better way to remove the brackets () first, before he disturbs the brackets {} which inclose the former, and then to remove these outer ones afterwards; thus, the last expression, upon removing the inner brackets, is 2 {a-3b+3c}: hence the expression, free of brackets, is 2a-6b+6c. SUBTRACTION. 24. Subtraction of algebra, like addition, can be actually performed only upon like quantities: if unlike quantities are to be subtracted the one from the other, the operation can only be indicated by placing the minus sign before the quantity to be taken away, inclosing this latter in brackets when it consists of several simple terms: that is, when it is a compound quantity. But agreeably to the principle already explained at (23), and illustrated in the examples above, a bracketed quantity, preceded by a minus or subtractive sign, may be written down free of brackets, provided the signs of all the terms which compose the subtractive quantity be changed. In other words, the operation of subtraction merely requires that we change the signs of all the terms of the subtractive quantity, and then collect the like terms as in addition. Thus, if 2a-3b+5c is to be subtracted from 7a+b-3c, the operation is indicated by 7a+b-3c-(2a-3b+5c)=7a+b-3c-2a+3b-5c= 5a+4b-8c, where the subtractive quantity, with changed signs, is added. The rule for subtraction is, therefore, simply this, namely: RULE. Change the signs of all the terms in the subtractive quantity, or conceive them to be changed, and then proceed as in addition: the result will be the correct remainder. (1) From 7a+ b—3c Or, Take 2a-3b+5c Rem. 5a+4b-8c Το 7a+b-3c Add -2a+3b-5c 5a+4b-8c (2) From 6a -5b+2c Or, To 6a-5b+2c Take 13a +76-2c Rem. 7a-12b+4c Add 13a7b+2c 7a-12b+4c And in this way is subtraction always convertible into addition: we have only to imagine every plus in the bottom or subtractive row, to be changed into a minus, and every minus into a plus, and then to add the two rows together. (3) 5ax2-3by2+4 2ax2+6by2+8 3ax2-9by2-4 (5) -4y+7x+6a -4x-y-4a (4) -8(a+x)-3(by) + ab -5 (a+x)+4(b-y)-2ab -3(a+x)—7(b—y)+3ab 3x-5y+2a (6) 7-3+ 4+1-2 9+5—6—3—4 -2-8+10+4+2 In this sixth example, we have used numbers only without letters, for the purpose chiefly of further illustrating the correctness of the operation. Thus, 7-3+4+1-2=7, and 9+5—6—3—4—1; and 1 subtracted from 7 leaves 6, which is the correct remainder; and the remainder above, namely, —2—8+10+4+2, is evidently 6 also. The learner may frame examples of this kind for himself, and then test the accuracy of the results as here shown:— (8) 4a√x-5b√y+a-b -6avy+3b√x+a+d (4a-3b)x+(6a-5b)√y-b-d In this tenth example, when the signs in the bottom row are changed, the coefficient of the second term will be —b; which, together with -a above, gives the sum (a+b): the coefficient of the third term, after the change, is b+c; which, together with the coefficient a-b, above, gives the sum a+c. (11) (a+b)(x-y)+(c-d)x2-(ab)y2+ab (a-b)(xy)-(c-d)x2+(a+b)y2+db 26(x-y)+2(c-d)x-2ay+(a-d)b (12) (a+b)x-(c+d)y+(e—f)z -(ab)x-(c-d)y+3fz NOTE. The terms addition and subtraction, as applied to the algebraic operations arranged under these heads, do not so clearly convey a correct idea of the processes so designated as the same terms do in common arithmetic. In the latter subject, a quantity added to another always increases it; and a quantity subtracted from another always diminishes it. Addition and subtraction, in the strict meaning of the word, must, indeed, always produce these effects. But, in algebra, this strict meaning is confessedly departed from; and the terms are used, in an extended sense, chiefly because no others more suitable, or more strictly descriptive of the operations intended offer themselves. In algebra, the addition, as it is called, of 6a and -4a is 2a; and the subtraction of -4a from 6a is 10a. The truth is, the term addition means, in algebra, the finding the balance of a set of quantities of which some may be marked for addition and some for subtraction: and what is called the sum tells us which class preponderates, and to what amount. Algebraic subtraction, as already shown, is converted into addition, by giving the opposite meaning to the additive and subtractive marks in the quantity to be taken away; and this is quite consonant with common sense: but a learner might easily be puzzled by questions and quibbles, expressed in common language, in reference to the operations here commented upon, because, as already remarked, the common acceptation of the terms addition and subtraction, differs from their meaning in algebra. We shall not perplex the beginner by entering upon any consideration of these quibbles. If he have clearly conceived the influence of the algebraic signs + and -, he knows that 6-(-4), that is, -4 subtracted (in the algebraic sense) from 6, gives 10, for 6-(-4) is the same as 6—(0-4); which implies that, if ◊ be taken from 6, the remainder will be too little by 4: but if 0 be taken away, the remainder, so to speak, will be 6; hence the correct remainder must be 10; or, to vary the illustration, 6 is the same as 6+4-4; and if -4 be taken from this, the remainder is 6+4 or 10. |