16. Addition of Similar Monomials.-(13.) 1. The sum of 7 a and 9 a is required. By 17, Law V, 7a+9a (7+9) a = 16 a. Hence, to add two positive similar terms, find the sum of their coefficients (11) and affix to the result the common letters. 2. Find the sum of 9 ab2c and 16 ab3c. By the rule above, 9 ab c+16 abc (9+16) ab3c = = 25 ab2c. 17. Addition of Polynomials of Plus Terms.-A polynomial is an algebraic expression of two or more terms. The addition of polynomials is accomplished by means of the second law of addition (II, 6 and 16, Rule). EXAMPLE. Find the sum of 6a9x, 3x+5a+6y3, and 2.r+a+mn. It is convenient in practice to write the expressions one underneath the other, with similar terms arranged in the same column. Find the sum of the terms in each column (16), and write the results connected with the plus sign. 5a3c × 11 ab3c2 = (5 aaac) (11abbbce) = 5 · 11 · aaaabbbccc = 55 a1b3c3 (by index, commutative, and associative laws of multiplication.) Similarly, the product of 9 ab'cmd" by 13 akb3cPdr is 9 ab'cmd"×13 ab3codTM — 9 · 13 · a·a2b1b3ç3çod"dr = 117 ak+1bl+3cm+pdn+r ̧ The coefficient 55 of the resulting product 55a3 is the product of the two coefficients 5 and 11 of the multiplicand 5 a3c and the multiplier 11ab3c; the literal part abc is the result of forming a product of all the different letters occurring in both multiplicand and multiplier, each with an exponent equal to the sum of the exponents of their letters in both multiplicand and multiplier. The product 117 ak+1bl+3cm+da+r is formed in a similar manner. The coefficient 117 is the product of the two coefficients 9 and 13, and the exponents k+1, +3, m+p, n+r are respectively the sums of the exponents of a, b, c, d in both the multiplicand and multiplier. From 7, laws III, IV for multiplication, and 18, the following rule for the product of two monomials is derived: To the product of the two monomial coefficients (11) annex the letters, each with an exponent equal to its exponent in the multiplicand plus its exponent in the multiplier. EXAMPLE.-Multiply 7 xyz by 3x3y5; m being a positive integer. 7xyz×3x3y=7·3x+3y1+5%= 21x+3y°z. By rule 19. The Multiplication of a Polynomial of Plus Terms by a Plus Monomial. From 27, Law V, a(b+c) = ab+ ac the following rule is derived: Multiply each term of the multiplicand by the multiplier, and add the partial products. EXAMPLE.-Multiply 2x3+5x+7 by 7 x2. By the rule above, (2x3+5+7) X (7 ) = (2 x3+5)7x +7(7) = (2 x3) (7 x2) + (5 x) (7 x2) +7(7 x2) The following exercises will serve as illustrations of the preceding definitions and rules of addition and multiplication. = EXERCISE I If a 1, b = 3, c = 4, d= 6, e = 2, ƒ = 0, x = 3, y=3, find the numerical values of the eight following algebraic expressions: and 1. Find the value of a +2b+4 c; here, a = 1, b=3, c=4, 2. ab+2bc+3ed. a+2b+4c=1+2·3+4·4 = 1 + 6 + 16 = 23. 3. ac+4 cd+3 cb. Add the following (see 16, 17): 9. 2x+5x+x+7, 3x2+2+6x3 +8x, x+3x3 +4, and 1+2x2+5x. 10. 2a+3b4d, 2b+3d+4c, 2d+3c+4a+4b, and 2c+3a. 11. x2+3 xy + y2+ x + y + 1,2x2+4 xy + 3y2+2x+2y+3, 3x2+5xy + 4y2+3x+4y+2, and 6x+10xy+5y2+x+y. 12. 2x2+ax2, x3+3 ax2, x3+2ax2 + a2x. 13. 4x+10a3+ax(5x+6a), 3(2a3+x3)+2ax(2x+a), x2(17x+19a) + 15 a2r, and 6a2 (x+3a) + a2 (7x+5a). (Remove parentheses by Law V, 87.) 14. 4 ab+x2, 3x2 +2 ab, 2 x (a+b) and 5 a (x+7b) + 11 x (x+13b+7a). Find the following products by 18, 19: 15. 3axy and 5ay*. 16. 2a, 3b, 4c, ab, bc, and abc. 17. Simplify (3 xy2) · (5 xy z3) · (4y2 zw) · (x y z w). 18. Find the product of: x+y+2 and 2.xyz 3x+5y+725 and 9xy223 3x+2a (x+2 ay+522) and 5 a2xy2z. CHAPTER II POSITIVE AND NEGATIVE NUMBERS ADDITION AND SUBTRACTION 20. In the conducting of business, capital is increased by gains and decreased by losses. Suppose that for a week's business the account stands: The total losses of $250.50 neutralize $250.50 of the gains or credits, leaving a net gain of $71.00. Suppose that for another week's business the above losses and gains were reversed: The credits, $250.50, are less than the losses, $321.50. Arithmetically the net loss is found by subtracting the credits, $250.50, from the total loss, $321 50, which gives $71.00. This result can be calculated algebraically as follows: Designate the credits plus (+) or positive dollars and the losses minus ( — ) or negative dollars, then the account can be represented: where the relation between + dollars and dollars is that, in balancing accounts, a given number of dollars cancels or neutralizes the same number of dollars; i. e., a loss of $75.00 neutralizes or cancels gains of $25.00 and $50.00. This result is expressed algebraically: = $25.00 +$50.00-$75.00 +$75.00-$75.00 = $0.00. The gains and losses are said to balance, or there is a balance of zero dollars, $0.00. Or if there should be a gain of $75.00 and two losses, one of $25.00 and one of $50.00, the result of the transactions would be expressed algebraically: +$75.00 – $25.00 - $50.00 = +$75.00 $75.00 $0.00. Zero is defined as the difference between two equal numbers. Suppose a loss of $321.50. It may be separated into two losses,. one of $250.50 and the other of $71.00, and the operation of balancing the dollars and dollars in the problem above would. + be indicated algebraically as follows: +$250.50. $321.50 + $250.50 = $0.00 - $71.00 $71.00. - $250.50 $71.00 That is, there remains a debt of $71.00, or the balance is a negative number. 21. The Series of Natural Numbers.-If from a fixed point 0, in a line A B, units of length are laid off to the right, the successivepoints so found can be designated by the series of natural numbers, 1, 2, 3, 4, 5, etc. A 1 2 3 4 5 6 7 8 9 10 11 12 0 B |