15. Find the product of 3 a, 4d, and 7 ad. ART. 31. What is the value of 3 a3 b2, if a = 2 and b=5? Putting these numbers instead of the letters, we have 3 a3 b23.23.52 3.2.2.2.5.5=600, Ans. Let the learner find the value of the following algebraical expressions, supposing a 1, b = 2, c = 3, and d=5. SECTION VIII. REDUCTION OF SIMILAR TERMS. ART. 32. Similar terms are those which are entirely alike with regard to the letters and exponents. N. B. Similarity of terms is wholly independent of the algebraic signs and, the numerical coefficients, and the order of the letters. Thus, a2b and 5 a2 b are similar; so also are 3 x2 y3 and 5 y3 x2. Any polynomial which contains similar terms, may always be reduced to a smaller number of terms. Of this the student has already had numerous examples in the reduction of equations. Let us take the polynomial 2 a2 b3 + 4 m3 n3 +7 a2 b3. m3 n3. The two terms 2 a2 b3 and 7 a2 b3 are similar, and it is evident that 7 times a quantity added to twice the same quantity, make 9 times that quantity; therefore, instead of 2 a2b3+7 a2 b3, we may write 9 a2 b3. Again, 4 m3 n3 and -m3 n3 are similar, and 4 times a quantity diminished by once the same quantity, leaves 3 times that quantity; hence, instead of 4 m3 n3 . m3 n3, we may put 3 m3 n3. By these reductions the polynomial 2 a2 b3+ 4 m3 n3 + 7 a2 b3 — m3 n3 becomes 9 a2 b3 +3 m3 n3. As a second example, take the polynomial 5 b2 c2. 2 m n2 + 17 b2 c2. - 10 m n2. Here, 5 b2c2+17 b2 c2 = 22 b2 c2, and - 2 m n2 -10 m n2 12 m n2; so that the given quantity, when reduced, becomes 22 b2 c2 - 12 m n2. For a third example, take 10 a2 63 +3 a3 b2 — 6 a2 b3+ 3 a2 b3 — 7 a3 b2 — 5 a2 b3 + 6 a3 b2. a3 62. First unite the terms of one kind, having the sign+; 10 a2 b3 +3 a2 b3 = 13a2b3; now unite the terms of the same kind, having the sign —; — 6 a2 b3 — 5 a2 b3 —— 11 a2 b3, hence, 10 a2 63+3 a2b3-6 a2 b3-5 a2 b3 13 a2 b3-11 a2 b3 =2a2b3. In like manner, 3 a3 62 +6 a3 b29 a3 b2, and -7a3 b2— a3 b2 — == -8 a3 b2; hence, instead of these four terms, we have 9 a3 b2-8 a3 b2, which is the same as a3 62 The given quantity, therefore, reduced, becomes 2 a2b3+ a3 b2. Whenever the total of the negative terms exceeds that of the similar positive terms, we take the difference of the two amounts, giving this difference the sign Suppose we have to reduce b+ 3 x2 y3 +7 x2 y3 — 10 x2 y3 — 5 x2 y3. Uniting the similar positive and negative terms separately, we have b+10 x2 y3 - 15 x2 y3 ; here we have 10 times x2 y3 added, and 15 times the same quantity subtracted, which is the same as subtracting 5 times that quantity; hence, instead of 10 x2 y3 - 15 x2 y3, we may put -5 x2 y3, and the original quantity becomes b — 5 x2 y3. We see that this last term might have been obtained by subtracting 10 from 15, and putting before the remainder the sign and after it the common letters with their proper exponents. ART. 33. From the preceding examples we derive the following RULE FOR THE REDUCTION OF SIMILAR TERMS. Unite all the similar terms of one kind affected with the sign+, by adding their coefficients, and writing the sum before the common literal quantity; unite, in like manner, the similar quantities of the same kind affected with the sign ; then take the difference between these two sums, and give to the result the sign of the greater quantity. Remark. To prevent errors, it is advisable to mark the terms as they are reduced. Let the learner reduce the following quantities. 1. a+3a+5a+6b+b+2a. 2. 10x-9y+17x-6x-2x+3y+10 y. 6. m2+10a2b3+24 a1b+2 a2b3-8a2b3-4 a1b4 a2b3-20 aa b. 7. 8abc+10x2y-4abc-10 x y2+3abc-4x2y +12 y2 x + abc. b3 8. 12 a2b3c10 a b3 c-6 a2 b3 c+8 a b3 c-2a3b c +4m-20 a b3 c. 9. 11 x2 y3 - 12 x2 y + 15 x y2 — 15 x2 y3 + a b c + 6x y2-4x2y-16x y2-2 abc. 10. x2+3x3 — 4 x2 + 9 x3 + 6 x2 y2 — 10 x2 —9 x3+ 5 x2 y2 + 15 x3 — 17 x2. 11. 3mn-6xy2+6xz-13 x y2-10 xz-2xy2+ 4xy2+2xz-11 y2 x. 12. 17 p2 q2+3abc-2pq-20 p2 q2 -3 a b c + 6pq+21 p2 q2-14pq+17. 13. 120+ m2 n3 — 6 x y + 15 m2 n2 — 24 x y + 5 — m3 n-16+4 m3 n. m3 14. 7x2y+3+4x-10 x2y-17-+3x-14x2y+ 5+2x+9+13x2 y +x y2. SECTION IX. ADDITION. ART. 34. The addition of positive monomials, as has already been seen, consists in writing them after each other, putting the sign + before each except that placed first, which is understood to be affected with the sign +. Thus, to add a, b, c, and d, we write a+b+c+d. ART. 35. The addition of the two polynomials, a+ b and c+d+e, in which all the terms are affected with the sign, gives a+b+c+d+e; for adding these two quantities is the same as adding the separate terms of which they are composed. As an example in numbers, let us add 10+3 and 6+9. Now, 10+3 is 13, and 6+9 is 15, and the sum of 13 and 15 is 28; also 10+3+6 +9 is 28. But if some terms of the quantities to be added have the sign these terms retain that sign in the sum. This is easily seen in the case of numbers. Let us add 9-3 to 13; 9 3 is 6, and 6 added to 13 gives 19. But if we first add 9 to 13, which is expressed thus, 13 +9, the sum is too great by 3, because we were to add only 9—3, or 6; we must, therefore, subtract 3 from 13+9, which gives for the true sum 13+9-3, or 19. Let us now add a b to m. Adding a to m, we have ma; this result is too great by b, because we were to add only ab, which is less than a by b; therefore, after having added a, we must subtract b, which gives for the correct result m+a— —b. We see, in the foregoing examples, that in adding we |