the sum of the positive and negative terms are respectively + 2x and -7x; and since the quantity which has to be subtracted exceeds that which has to be added by 5x, 5x has therefore to be written down with a negative sign prefixed. In the third column the sum of the positive terms is 5y, and that of the negative terms is y, the result therefore (from what has been said above) is +4y: therefore the whole sum is 11x3-5x+4y. 5. Add together a3+3a2b+3ab2+b3, a3-3a2b+3ab2-b3, -a3+3a2b -3ab2+b3, —a3-3a2b-3ab2-b3, and a3+6a2b+6ab2+b3. 6. Add together a+b+c+d, a+b+c+d, a−b+c+d, a+b-c+d, and a+b+c-d. 7. Add together x+3ax3-bx2+3cx-d, 4x2-6αx3+5bx2-3cx+2d, -54-6ax3-5bx2-3cx-2d, 6x+7ax3-8bx2+6cx + 7d, - 10x-ax3 +10bx2-cx+d, and -x-ax3-bx2-cx-d. 8. Find the sum of a5+5a4b+6a2b2c-7ab, 6a5-a1b-6a2b2c+10ab, -2a5+4a1b+12a2b2c-10ab, 5a5-16a1b-11a2b2c+13ab, and -10a3 +8a1b + a2b2c-6ab. 9. Find the sum of am-2b+3c-4d, 23a+16b"-13c2+12d1, -14am +15b"-17c"+ 19d, 18am- b" + co-d, and -13a+5b"- 3cr 10. x6-6ax+3a2x2-28a3x3+9a4x2-54α5x+27a -3x6+7ax- 3⁄4a2x2- a3x3+3a2x2+27a5x- as a2x 3 4 1 3 5 11. Find the sum of a3- a2x + 1⁄2 ax2 — 1⁄2 x23‚ a3 + 1; a2x+ 4 24 8 2 12. Add together 5a2b-7a3bc-13b2c1+10, a2b+8a3bc−10b2c1+15 219 CASE III. Where the quantities are unlike. RULE. "Set all the quantities down one after another with their coefficients and proper signs prefixed, and collect all the like quantities together (if there be any) by the foregoing Rules." Ex. Add together 7a-3b+5c-10d, 2b-3c+d-4e, 5c-6a-4e+2d, -3b-8c+7a-e, and 21e-16c+a-5d. Proceeding by the above Rule, first arranging like quantities under each other, The Reason for the above process appears from what has been said in Cases I. and II. 2. Add together 2a3+4b2x-c2x2, 2c2x2+4a3—6b2x, and 262x-4c2x2+a3. 3. Find the sum of 3ab2-4a2b+a3, — 4ac2+5ab2 — c3, −7b3 +2a2b—6ac2, 5a3-11ab2-12ac2. 4. Add together 3x3— x2y+xy2—3y3, 6xy2−2y3+12x2y— x3, 4y3-4x3-2x2y, 7x2y-7y3, -9x3+9xy2-12y3+2x2y, and 5xy3-5x2y. 5. Add together 5a3b+12a2bc-abc2+2c3-4b2c2+15, a2+11abc2 +6-20a3b, c3-a2bc+5b3+17+12a4, 53abc2-4b3+5b2c2a2-2abc+5c3, and 4a2bc+13-7a3+14a3h; and express the result in numbers, if a=0, b=1, c=1. 3ab2-4a3+2bc2 6. Find the sum of a1-3ab2-3ac2 + b2c2+3c1, +6ac2-b1, 3a3+2a2c2-2b2c2+3ca3-5ac2, and 464-6a2b-3a1-3c1+b2c2. 7. Find the coefficient of x2y in the sum of 5x3+3x2y+4xy2—y3, 6y3 + xy2 — 5x2y—7x3, x3—5xу2+x2y-6y3, −2x2y+3xy3+x3. 8. Add together a-3b+1c-3d, −1c+}a−1b+d, 1d-}b+c-a, a-3d+b-gc, and 8a-6b+3c-4d. 9. Find the coefficient of b in the expression 24. ર SUBTRACTION. RULE. Change the signs of all the quantities to be subtracted, or conceive them to be changed, and then collect the different terms together as in the cases of Addition." Ex. Subtract b-c from a. Proceeding by the Rule given above a b-c Reason for the process. If 6-3 or 3 be subtracted from 10, it is evident that the remainder will be greater by 3, than if 6 were subtracted. For the same reason, if b-c be subtracted from a, the remainder will be greater by c, than if b were subtracted. Now if b be subtracted from a, the remainder =a-b; therefore if b-c be subtracted from a the remainder is a-b+c. Hence the general rule given above. It is here assumed that a>b and b>c, in this case the rule is capable of proof as above; in other cases it is assumed to be true. 7. From a1-4a3b+6a2b2-4ab3+b4 take a1+4a3b+6a2b2+4ab3+ba. 8. From 2x+11a+10b-5c-23 take 2c-10+5a-3b. 9. From +3αx3-2bx2+3cx-4d take 3αx+ax3-4x2+6cx+d. 10. From 724-78x3y-10x3y2+17xy3+3y^ take +36x3y+10x3y2—17xy3 +34y1. take 3x2-4y2+5xz−6x2+7yz-8xy; 11. From 6x2+7xy-5y2-12xz-8yz-5x2 and find the numerical value of the difference when x=1, y=0, x=3. 12. From -a2x2+4axy-3aby take 4aby2-5axy+2a2x2. 13. From ax3+bx2+cx+d take ex3-bx2+ cx+4d. 14. From ax + bx2y-cxy2-dy3 take ey3+fxy-gx3y-hx3. 15. From 5a2xy-4bxy2+9ax2y+15y1 take 3a2xy+5b2x2y—9cxy2 — 5y1. 16. From 10am-15b"-c+5d take - 9am +2b" + c2 — 5do. 19. Subtract 2a+36+1c+1bc from 5a-76-3bc+1c. 20. Subtract 3y+1a-3x from 1y-za-2x+3a. 21. From the sum of the first four of the following expressions, a2+b2+c2+d2, d2+b2+c2, a2-c2+b2-d3, a2-b2+c2+ď2, b2+c2+d2-a3, subtract the sum of the last four. 25. On the removal of brackets or vincula from algebraical expressions. Since the sign + before a bracket or vinculum implies that all quantities within the bracket or under the vinculum are to be added to whatever quantities go before, and since the sign - before a bracket or vinculum implies that all quantities within the bracket or under the vinculum are to be subtracted from whatever quantities go before; it is clear that a bracket or a vinculum, with a positive sign before it, can be removed from an algebraical expression without altering the signs of the different terms within the bracket or under the vinculum; and that a bracket or a vinculum with a negative sign before it can be removed from an algebraical expression, if the signs of all the terms within the bracket or under the vinculum be changed. So also in the case of a double bracket, a+{b-(c+d)}=a+b−(c+d)=a+b-c-d, Ex. IX. A. Expressions, which involve single brackets, reduced to their most simple forms: 1. 6a-4b-(5a+2b)=6a-4b-5a-2b-a-6b. 2. (3a-x)-(4x-2a)+(3-x)-(a−4x)+(x-1) =3a-x-4x+2a+3-x-a+4x+x-1=4a−x+2. 3. 2a2-(-3a2+5b2-c2)+(5b2-c2)=2a2+3a2-562+ c2+5b3 — c2=5a3. Examples for Practice. Ex. IX. A. Reduce to their most simple forms the following expressions: (4) a3-a3b+ab3 — b3 — (a3 +b3) + (a2b—ab2). |