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Unit 13

Unit 13. BASIC ALGEBRAIC OPERATIONS. ADDITION. Only like terms can be added. The addition of unlike terms can only be indicated Procedure for adding like terms: Add the numerical coefficients, applying the procedure for addition of signed numbers Leave the variables unchanged

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Unit 13

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  1. Unit 13 BASIC ALGEBRAIC OPERATIONS

  2. ADDITION • Only like terms can be added. The addition of unlike terms can only be indicated • Procedure for adding like terms: • Add the numerical coefficients, applying the procedure for addition of signed numbers • Leave the variables unchanged 6y + (–5y) = 1y = yAns –13ab + (–11ab) = –24abAns

  3. ADDITION • Procedure for adding expressions that consist of two or more terms: • Group like terms in the same column • Add like terms and indicate the addition of the unlike terms • Add: 5y + (–3x) + 6x2y and (–4x) + (–2y) + (–2x2y) • Group like terms in the same column • Add the like terms and indicate the addition of the unlike terms

  4. SUBTRACTION • Just as in addition, only like terms can be subtracted • Each term of the subtrahend is subtracted following the procedure for subtraction of signed numbers • Subtract: (7x2 + 7xy – 15y2) – (–8x2 + 5xy – 10y2)Change the sign of each term in the subtrahend and follow the procedure for addition of signed numbers

  5. MULTIPLICATION • In multiplication, the exponents of the literal factors do not have to be the same to multiply the values • Procedure for multiplying two or more terms: • Multiply the numerical coefficients, following the procedure for multiplication of signed numbers • Add the exponents of the same literal factors • Show the product as a combination of all numerical and literal factors • Multiply: (–4)(5x)(–6x2y)(7xy)(–2y3) • Multiply all coefficients and add exponents of the same literal factors = (–4)(5)(–6)(7)(–2)(x1 + 2 + 1)(y1 + 1 + 3) = –1680x4y5 Ans

  6. MULTIPLICATION • Procedure for multiplying expressions that consist of more than one term within an expression: • Multiply each term of one expression by each term of the other expression • Combine like terms b. (2a – 3b)(5a + 2b) = (2a)(5a) + (2a)(2b) + (–3b)(5a) + (–3b)(2b) a. 2x(3x2 + 2x – 5) = 2x(3x2) + 2x(2x) + (2x)(–5) = 10a2 + 4ab – 15ab – 6b2 = 6x3 + 4x2 – 10xAns = 10a2 – 11ab – 6b2 Ans

  7. DIVISION • Procedure for dividing two terms: • Divide the numerical coefficients following the procedure for division of signed numbers • Subtract the exponents of the literal factors of the divisor from the exponents of the same letter factors of the dividend • Combine numerical and literal factors

  8. DIVISION • Divide: (40a3b4c5)  (–4ab2c3) = (–10)(a3 – 1)(b4 – 2)(c5 – 3) = –10a2b2c2 Ans

  9. DIVISION • Procedure for dividing when the dividend consists of more than one term: • Divide each term of the dividend by the divisor, following the procedure for division of signed numbers • Combine terms • Divide:

  10. POWERS • Procedure for raising a single term to a power: • Raise the numerical coefficients to the indicated power following the procedure for powers of signed numbers • Multiply each of the literal factor exponents by the exponent of the power to which it is raised • Combine numerical and literal factors • Procedure for raising two or more terms to a power: • Apply the procedure for multiplying expressions that consist of more than one term (FOIL) (2x2y3)2 = 22(x2)2(y3)2 = 4x4y6 Ans (x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2 Ans

  11. ROOTS • Procedures for extracting the root of a term: • Determine the root of the numerical coefficient following the procedure for roots of signed numbers • The roots of the literal factors are determined by dividing the exponent of each literal factor by the index of the root • Combine the numerical and literal factors • Solve: = –2a2b3cAns

  12. REMOVAL OF PARENTHESES • Procedure for removal of parentheses preceded by a plus sign: • Remove the parentheses without changing the signs of any terms within the parentheses • Combine like terms 7x + (–4x + 3y – 2) = –7x – 4x + 3y – 2 = –11x + 3y – 2 Ans • Procedure for removal of parentheses preceded by a minus sign: • Remove the parentheses and change the sign of each term within the parentheses • Combine like terms 9a – (–4a + 2b – 6) = 9a + 4a – 2b + 6 = 13a – 2b + 6 Ans

  13. COMBINED OPERATIONS • Expressions that consist of two or more different operations are solved by applying the proper order of operations • Simplify: 15x – 4(–2x) + x = 15x + 8x + x = 23x + x = 24xAns • Simplify: [3x – x + (x2y3)2]2 = [3x – x + x4y6]2 = [2x + x4y6]2 = (2x + x4y6)(2x + x4y6) = 4x2 + 2x5y6 + 2x5y6 + x8y12 = 4x2 + 4x5y6 + x8y12 Ans

  14. BINARY NUMERATION SYSTEM • The binary number system uses only the two digits 0 and 1. These two digits are the building blocks for the binary code that is used to represent data and program instructions for computers • Place values for binary numbers are shown below: Remember this can continue as far as needed in either direction

  15. EXPRESSING BINARY NUMBERS AS DECIMAL NUMBERS • We use a subscript 2 to show that a number is binary • Express the binary number 1012 as a base 10 decimal number: 1012 = 1(22) + 0(21) + 1(20) = 4 + 0 + 1 = 510Ans • Express the decimal number 1910 as a binary number: • The largest power of two that will divide into 19 is 24 or 16 19 = 1(24) + 0(23) + 0(22) + 1(21) + 1(20) = 100112Ans

  16. PRACTICE PROBLEMS • Perform the indicated operations and simplify: • 6a + 7a + 9b • (–3xy) + 4x + (–5xy2) + 5xy + (–7x) • –3.07ab + 7.69c + (–5.76ab) + 9d + (–11.2c) • 1/2x + (–2/3y) + 1/4z + (–1/3z) + 2/3x • 4x2y + (–5xy2) + 7xy2 + (–2x2y) • 7a – 3a • –10x – (–20x) • (3y2 – 4z) – (–2y2 + 4z) • –1 1/2ab – 1 2/3ab • (–2.04t2 + 7.6t – 7) – (3t2 – 6.7t – 4)

  17. PRACTICE PROBLEMS (Cont) • (3ab)(–4a2b2) • (–1/2x)(–1/3y2)(1/4x3) • (a – b)(a – b) • (2x2 – 3y)(–3x2 + 2y) • 16y2 4y • 1 1/3 a2b3  2/3ab • (2.4x3y3 + 4.8x2y2– 24x)  1.2x • (x2y)3 • (–2.1a2b3)2 • (2/3 x3y3z2)3 • (–2m2n + p3)2

  18. PRACTICE PROBLEMS (Cont) 22. 23. 24. 25. 2x – (x – 2y) 26. –(x + y – z) – (x2 – y2 + z) 27. 15 – (ab – a2b – b) – 4 + (ab – b) 28. (18a4b2)  (3a2b) – b3(b2) 29. 5(2x – y)2 – (x2 – y2)

  19. PROBLEM ANSWER KEY • 13a + 9b • 2xy + (–3x) + (–5xy2) • –8.83ab + (–3.51c) + 9d • 7/6x + (–2/3y) + (–1/12z) • 2x2y + 2xy2 • 4a • 10x • 5y2 – 8z • –3 1/6ab • –5.04t2 + 14.3t – 3

  20. PROBLEM ANSWER KEY (Cont) • –12a3b3 • 1/24x4y2 • a2 – 2ab + b2 • –6x4 + 13x2y – 6y2 • 4y • 2ab2 • 2x2y3 + 4xy2 – 20 • x6y3 • 4.41a4b6 • 8/27x9y9z6 • 4m4n2 – 4m2np3 + p6

  21. PROBLEM ANSWER KEY (Cont) • 11a3b2c • – 4x4y3 • x + 2y • –x2 + y2 – x – y • 11 + a2b • 6a2b – b5 • 19x2– 20xy + 6y2

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