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4.2 Algebraic Properties: Combining Expressions We begin this section with a summary of the algebraic properties of numbers. Property Name Property Example Commutative property a+b=b+a !9 + 4 = 4 + (!9) (of addition) Commutative property a•b = b•a 5 • (!12) = !12 • 5 (of multiplication) Associative property a + (b + c) = (a + b) + c 6 + ( 5 + 3) = ( 6 + 5 ) + 3 (of addition) Associative property a • (b • c) = (a • b) • c !3 • ( 4 • 5 ) = (!3 • 4) • 5 (of multiplication) Identity property a+0=0+a=a !9 + 0 = 0 + (!9) = !9 (of addition) Identity property a •1 = 1• a = a !13 • 1 = 1 • (!13) = !13 (of multiplication) Inverse property x + (!x) = 0 7 + (!7) = 0 (of addition) Inverse property 1 1 If x " 0, x • = 1 5• =1 (of multiplication) x 5 Distributive property a ( b + c ) = ab + ac 6 ( x + 3) = 6x + 18 (addition form) Distributive property a ( b ! c ) = ab ! ac 4 ( x ! 5 ) = 4x ! 20 (subtraction form) Multiplication property of 0 a•0 = 0•a = 0 !12 • 0 = 0 • (!12) = 0 Zero factor property If a • b = 0, a = 0 or b = 0 If ! 7 • y = 0, y = 0 The two inverse properties listed here are new. Our first example illustrates the use of the commutative and associative properties applied to algebraic expressions. 281 Example 1 Apply the commutative and/or associative properties to rewrite and simplify each expression. a. b. c. d. Solution a. 5x • 7 4 + ( 7 + 3x ) 8y • 5y 3 12x • 4 Applying the commutative property (of multiplication): 5x • 7 = 7 • 5x commutative property = 35x b. multiplication Applying the associative and commutative properties (of addition): 4 + ( 7 + 3x ) = ( 4 + 7 ) + 3x associative property = 11 + 3x = 3x + 11 c. Applying the commutative property (of multiplication): 8y • 5y = 8 • 5 • y • y commutative property = 40y 2 d. addition commutative property multiplication Applying the commutative and associative properties (of multiplication): 3 3 12x • = • 12x commutative property 4 4 !3 $ = # • 12 & x associative property "4 % = 9x multiplication One major topic involved with combining algebraic expressions is that of like terms. Suppose we are given the expressions 6x + 8x and 282 5x 2 ! 12x 2 Is there a way to combine them? The answer lies in the use of the distributive property. Writing the distributive properties (in reverse) as: ab + ac = a ( b + c ) = ( b + c ) a ab ! ac = a ( b ! c ) = ( b ! c ) a Note that these properties allow us to merely add (or subtract) the numbers in front of the variables (these numbers are called the coefficients of the term). Applying the distributive property to these two expressions: 6x + 8x = ( 6 + 8 ) x = 14x 5x 2 ! 12x 2 = ( 5 ! 12 ) x 2 = !7x 2 Note that, in order for two expressions (or terms) to be considered like terms, they must have exactly the same variables and exponents on those variables. Thus 5a 2b 3 and !14a 2b 3 are like terms (their variables and exponents are identical), while 7x 3 y 2 and 12x 3 y 3 are not like terms (the exponents on the y variable are not identical). The first two terms can be added, while the second two terms cannot be added. Example 2 Use the idea of like terms to combine the terms. a. b. c. d. Solution 5x ! 18x 13xy 2 + 29xy 2 27x 3 y 2 ! 6x 3 y 2 ! 12x 3 y 2 12a 2b 2 + 9a 2b 2 ! 7a 3b 2 a. The two terms are like terms. Using the distributive property: 5x ! 18x = ( 5 ! 18 ) x = !13x b. The two terms are like terms. Using the distributive property: 13xy 2 + 29xy 2 = (13 + 29 ) xy 2 = 42xy 2 c. All three terms are like terms. Using the distributive property: 27x 3 y 2 ! 6x 3 y 2 ! 12x 3 y 2 = ( 27 ! 6 ! 12 ) x 3 y 2 = 9x 3 y 2 283 d. The first two terms are like terms, but the third term is not (the exponent on the a variable is different). Using the distributive property with the first two terms: 12a 2b 2 + 9a 2b 2 ! 7a 3b 2 = (12 + 9 ) a 2b 2 ! 7a 3b 2 = 21a 2b 2 ! 7a 3b 2 We can combine the properties of numbers, together with the idea of like terms, to combine more algebraic expressions. Consider the expression: 2x 2 + 3x + 5 ! 5x 2 ! 4x + 9 The commutative property (of addition) allows us to reorganize the terms with like terms adjacent to each other: 2x 2 + 3x + 5 ! 5x 2 ! 4x + 9 = 2x 2 ! 5x 2 + 3x ! 4x + 5 + 9 Now use the distributive property to combine like terms: 2x 2 + 3x + 5 ! 5x 2 ! 4x + 9 = 2x 2 ! 5x 2 + 3x ! 4x + 5 + 9 = (2 ! 5) x2 + ( 3 ! 4 ) x + (5 + 9) = (!3)x 2 + (!1)x + 14 = !3x 2 ! x + 14 Note how parentheses were dropped, and addition changed to subtraction, in the last step. Example 3 Simplify the following expressions. a. b. c. d. 5x 2 + 2x ! 9 ! 7x 2 ! 8x + 4 5x + 7xy ! 3y ! 9x ! 15xy 7a 2 ! 5ab + 2b 2 ! 5a 2 ! 12ab ! 8b 2 1 3 2 5 1 x + xy + y ! x + xy ! 3y 2 4 3 6 3 284 Solution a. Use the commutative property to rearrange terms, then the distributive property to combine like terms: 5x 2 + 2x ! 9 ! 7x 2 ! 8x + 4 = 5x 2 ! 7x 2 + 2x ! 8x ! 9 + 4 = ( 5 ! 7 ) x 2 + ( 2 ! 8 ) x + ( !9 + 4 ) = (!2)x 2 + (!6)x + (!5) = !2x 2 ! 6x ! 5 b. Use the commutative property to rearrange terms, then the distributive property to combine like terms: 5x + 7xy ! 3y ! 9x ! 15xy = 5x ! 9x + 7xy ! 15xy ! 3y = ( 5 ! 9 ) x + ( 7 ! 15 ) xy ! 3y = (!4)x + (!8)xy ! 3y = !4x ! 8xy ! 3y c. Use the commutative property to rearrange terms, then the distributive property to combine like terms: 7a 2 ! 5ab + 2b 2 ! 5a 2 ! 12ab ! 8b 2 = 7a 2 ! 5a 2 ! 5ab ! 12ab + 2b 2 ! 8b 2 = ( 7 ! 5 ) a 2 + ( !5 ! 12 ) ab + ( 2 ! 8 ) b 2 = 2a 2 + (!17)ab + (!6)b 2 = 2a 2 ! 17ab ! 6b 2 d. Use the commutative property to rearrange terms, then the distributive property to combine like terms: 1 3 2 5 1 x + xy + y ! x + xy ! 3y 2 4 3 6 3 1 5 3 1 2 = x ! x + xy + xy + y ! 3y 2 6 4 3 3 " 1 5% " 3 1% "2 % = $ ! ' x + $ + ' xy + $ ! 3' y # 2 6& # 4 3& #3 & 4% " 3 5% " 9 " 2 9% = $ ! ' x + $ + ' xy + $ ! ' y # 6 6& # 12 12 & # 3 3& 1 13 7 = ! x + xy ! y 3 12 3 285 The distributive property plays another role (other than combining like terms) in simplifying expressions. Consider the expression: ( 5 x 2 ! 3x + 6 ) We can apply the distributive property to multiply this expression: ( ) 5 x 2 ! 3x + 6 = 5 • x 2 ! 5 • 3x + 5 • 6 = 5x 2 ! 15x + 30 We consider this last expression simplified, since it is easier to combine it with other expressions. Example 4 Simplify the following expressions by applying the distributive property. a. ) 6 2a 2 ! 5a ! 9 b. !4 3y ! 8y + 12 c. 5 3% "2 12 $ x 2 ! x + ' #3 6 8& d. Solution ( a. ( 2 ) 4x ( 3x ! 4y ! 7 ) Applying the distributive property: 6 2a 2 ! 5a ! 9 = 6 • 2a 2 ! 6 • 5a ! 6 • 9 ( ) = 12a 2 ! 30a ! 54 b. Applying the distributive property: !4 3y 2 ! 8y + 12 = (!4) • 3y 2 ! (!4) • 8y + (!4) • 12 ( ) = !12y 2 ! (!32y) + (!48) = !12y 2 + 32y ! 48 c. Applying the distributive property: 5 3% 2 5 3 "2 12 $ x 2 ! x + ' = 12 • x 2 ! 12 • x + 12 • #3 6 8& 3 6 8 9 = 8x 2 ! 10x + 2 286 d. Applying the distributive property: 4x ( 3x ! 4y ! 7 ) = 4x • 3x ! 4x • 4y ! 4x • 7 = ( 4 • 3) x 2 ! ( 4 • 4 ) xy ! ( 4 • 7 ) x = 12x 2 ! 16xy ! 28x Our final type of problem in this section combines the use of the distributive property and combining like terms. Consider the expression: ( ) ( 3 x 2 ! 5x ! 7 ! 2 x 2 ! 6x + 4 ) We can apply the distributive property to each parenthesis to obtain: ( ) ( 3 x 2 ! 5x ! 7 ! 2 x 2 ! 6x + 4 ) = 3 • x 2 ! 3 • 5x ! 3 • 7 + (!2) • x 2 ! (!2) • 6x + (!2) • 4 ( ) = 3x 2 ! 15x ! 21 + !2x 2 ! (!12x) + (!8) = 3x 2 ! 15x ! 21 ! 2x 2 + 12x ! 8 Note that this final expression has like terms which can be combined. Continuing to combine like terms: ( ) ( 3 x 2 ! 5x ! 7 ! 2 x 2 ! 6x + 4 ) = 3x 2 ! 15x ! 21 ! 2x 2 + 12x ! 8 = 3x 2 ! 2x 2 ! 15x + 12x ! 21 ! 8 = ( 3 ! 2 ) x 2 + ( !15 + 12 ) x + ( !21 ! 8 ) = 1x 2 + (!3x) + (!29) = x 2 ! 3x ! 29 Note how we used the identity property (of multiplication) to write 1x 2 as x 2 in this last step. 287 Example 5 Simplify the following expressions. a. 5 ( 3x ! 4y ) ! 7 ( 7x ! 6y ) b. !4 x 2 ! 4x ! 5 3x 2 ! 5x c. d. Solution a. ( ) ( ) 6 ( y ! 5y ! 6 ) ! 4 ( 2y ! 3y ! 1) 2 2 1 $ 1 $ !1 !1 3 # x + y& ' 4 # x ' y& "2 "3 4 % 6 % Applying the distributive property and combining like terms: 5 ( 3x ! 4y ) ! 7 ( 7x ! 6y ) = 5 • 3x ! 5 • 4y + (!7) • 7x ! (!7) • 6y = 15x ! 20y + (!49x) ! (!42y) = 15x ! 20y ! 49x + 42y = 15x ! 49x ! 20y + 42y = (15 ! 49 ) x + ( !20 + 42 ) y = !34x + 22y b. Applying the distributive property and combining like terms: !4 x 2 ! 4x ! 5 3x 2 ! 5x ( ) ( ) = (!4) • x 2 ! (!4) • 4x + (!5) • 3x 2 ! (!5) • 5x ( ) = !4x 2 ! ( !16x ) + !15x 2 ! ( !25x ) = !4x 2 + 16x ! 15x 2 + 25x = !4x 2 ! 15x 2 + 16x + 25x = ( !4 ! 15 ) x 2 + (16 + 25 ) x = !19x 2 + 41x 288 c. Applying the distributive property and combining like terms: 6 y 2 ! 5y ! 6 ! 4 2y 2 ! 3y ! 1 ( ) ( ) = 6 • y 2 ! 6 • 5y ! 6 • 6 + (!4) • 2y 2 ! (!4) • 3y ! (!4) • 1 ( ) = 6y 2 ! 30y ! 36 + !8y 2 ! ( !12y ) ! (!4) = 6y 2 ! 30y ! 36 ! 8y 2 + 12y + 4 = 6y 2 ! 8y 2 ! 30y + 12y ! 36 + 4 = ( 6 ! 8 ) y 2 + ( !30 + 12 ) y + ( !36 + 4 ) = !2y 2 + ( !18y ) + (!32) = !2y 2 ! 18y ! 32 d. Applying the distributive property and combining like terms: 1 $ 1 $ !1 !1 3 # x + y& ' 4 # x ' y& "2 "3 4 % 6 % 1 1 1 1 x + 3 • y + ('4) • x ' ('4) • y 2 4 3 6 3 3 ! 4 $ ! 2 $ = x + y + # ' x & ' # ' y& " 3 % " 3 % 2 4 3 3 4 2 = x+ y' x+ y 2 4 3 3 3 4 3 2 = x' x+ y+ y 2 3 4 3 ! 3 4$ ! 3 2$ =# ' & x+# + & y " 2 3% " 4 3% = 3• 8$ ! 9 8$ ! 9 =# ' & x+# + & y " 6 6% " 12 12 % = 1 17 x+ y 6 12 289 One final note regarding the opposite of an expression. Given the expression: ( ! x 2 ! 4x ! 7 ) Because of the identity property (of multiplication), we can write the expression as: ( ) ( ! x 2 ! 4x ! 7 = !1 x 2 ! 4x ! 7 ) Now applying the distributive property and simplifying: ( ) ( ! x 2 ! 4x ! 7 = !1 x 2 ! 4x ! 7 ) = !1 • x ! (!1) • 4x ! (!1) • 7 2 = !x 2 ! ( !4x ) ! ( !7 ) = !x 2 + 4x + 7 Note that the opposite of the expression is a complete reversal of all signs. Though it is useful to remember this, inserting the 1 and applying the distributive property will produce the same result. Terminology properties of numbers coefficient like terms Exercise Set 4.2 Apply the commutative and/or associative properties to rewrite and simplify each expression. 1. 3. 5. 7. 9. 11. 13. 15. 3x • 9 !8x • 9 !7x • (!13) 5 + ( 9 + 8x ) !15 + ( 4 ! 11x ) !8 + (13x ! 6 ) 7x • 9x !6y • 8y 2. 4. 6. 8. 10. 12. 14. 16. 290 7x • 12 !6x • 13 !12x • (!11) 13 + (12 + 15x ) !13 + ( !5 ! 14x ) !12 + ( !7x + 5 ) 12x • 13x !14a • ( !15a ) 17. 12y • 2 3 18. 16y • 4 5 2 4 21. ! a • a 3 5 2 " 5 % 23. $ ! x ' # 6 & 3 4 7 10 5 3 22. ! x • x 6 10 2 " 3 % 24. $ ! b ' # 4 & 19. !25y • 20. !30y • Give the full name of the property illustrated by each statement. 25. 4x + ( 3x + 2 ) = ( 4x + 3x ) + 2 27. 1 • 9x = 9x 26. 9 + 8x = 8x + 9 28. 5 + ( 8 ! 5x ) = 13 ! 5x 3 " 4% 30. ! • $ ! ' = 1 4 # 3& 32. 143x • 0 = 0 34. ! ( 3 ! 4x ) = !1( 3 ! 4x ) 36. !9 ( y + 5 ) = !9y ! 45 38. !9y • 6 = !54y 29. 5x • 9 = 45x 31. 33. 35. 37. If ! 5x = 0, then x = 0 . 3 ! 5y = !5y + 3 12x + ( !12x ) = 0 5 ( x ! 8 ) = 5x ! 40 39. !59y • 0 = 0 9 " 10x % •$! 41. ! ' =1 10x # 9 & 43. 3( 8x ) = 24x 45. 0 ! 14y = !14y 40. If x 2 = 0, then x = 0 . 42. 8y + ( !8y ) = 0 44. !12x + 0 = !12x 46. !5 (12y ) = !60y Use the idea of like terms to combine the terms. 47. 49. 51. 53. 55. 7x ! 15x !8x 2 ! 5x 2 23x 2 y 3 ! 47x 2 y 3 48. 50. 52. 54. 56. 13x 3 y 2 ! 18x 3 y 2 ! 15x 3 y 2 !4a 3b 2 + 8a 2b 3 ! 7a 3b 2 291 12x ! 45x !12x 3 ! 9x 3 !34x 3 y 4 ! 25x 3 y 4 !12x 3 y 2 ! 28x 3 y 2 + 7x 3 y 2 !13a 2b 3 + 5a 2b 3 ! 8a 3b 2 Simplify the following expressions. 57. 59. 61. 63. 65. 67. 69. 71. 73. 75. 76. 3x 2 + 4x ! 5 ! 5x 2 ! 9x + 12 58. 2 2 3x + 11x ! 14 ! 7x ! 7x + 7 60. 4a ! 6ab ! 5b ! 11a ! 9ab 62. !8a ! 7ab + 5b ! 3a ! 9b 64. 2 2 2 2 5u ! 7uv + 3v ! 5u ! 14uv ! 9v 66. 2 2 2 2 !3u ! 7uv + 6v + 12u ! 9uv ! 4v 68. 70. 0.13x ! 0.47y ! 0.67x ! 2.9y 3 2 4 5 72. x! y! x! y 4 3 5 6 1 1 1 1 1 74. x + xy + y ! x ! xy ! 2y 4 3 2 6 2 5.6x 2 ! 0.87xy ! 6.8y 2 ! 3.9xy ! 5.3y 2 !2.9x 2 ! 3.68xy ! 5.78y 2 ! 6.5xy + 6.7y 2 5x 2 + 8x ! 9 ! 12x 2 ! 4x + 4 !5x 2 ! 13x + 2 ! 7x 2 + 4x ! 9 !5a ! 6ab + 4b ! 8a + 12ab !7a ! 12ab + 6b ! 13a ! 4b 4u 2 ! 9uv ! 2v 2 ! 7u 2 ! 13uv + 12v 2 !13u 2 ! 6uv + 4v 2 + 5u 2 ! 7uv ! 4v 2 3.57x ! 2.56y ! 6.7x ! 5.8y 2 1 2 1 x! y! x+ y 7 4 3 3 1 2 3 2 1 ! x + xy + y ! x ! 2xy ! y 2 3 4 3 6 Simplify the following expressions by applying the distributive property. ( 77. 4 3x 2 ! 4x ! 12 ( ) 79. !5 6a 2 + 12a ! 15 ) 2 5% "3 81. 12 $ x 2 ! x + ' #4 3 6& 1 3% "2 83. !9 $ y 2 ! y + ' #3 3 4& ( 85. 4 5.3x 2 ! 6.7x ! 9.6 ( ( 89. 5x ( 4x ! 8y ! 9 ) 91. !6y ( 4x ! 5y ! 9 ) ( 80. !9 !4a 2 + 13a ! 22 ) 87. !5 8.6x ! 5.9x + 8.46 2 ) 78. 7 5x 2 ! 6x ! 13 ) 1 4% "3 82. 15 $ x 2 ! x + ' #5 3 9& 1 1% "1 84. !10 $ y 2 ! y + ' #3 4 2& ( 86. 8 3.7x 2 ! 5.6x ! 8.4 ) ( ) 88. !8 3.4x + 6.7x ! 9.5 2 90. 12x ( 8x ! 9y ! 13) 92. !12y ( 8x ! 13y ! 11) ) Simplify the following expressions. 93. 3( 4x ! 5y ) ! 5 ( 6x ! 7y ) 95. !4 ( 5a ! 6b ) ! 7 ( !3a + 2b ) 94. 6 ( 3x ! 7y ) ! 7 ( 5x ! 8y ) 96. !5 ( 6a ! 9b ) ! 8 ( !4a + 9b ) 97. 5 x 2 ! 6x ! 8 2x 2 ! 5x 98. !4 3x 2 ! 4x ! 5 5x 2 ! 4x ( ) ( ) ( 292 ) ( ) ( ) ( ) 101. 4 ( y ! 4y ! 5 ) ! 6 ( 2y ! 5y ! 3) 103. !4 ( 2x ! 3x + 5 ) ! 7 ( !2x + 3x ! 7 ) 99. !6 2y 2 ! 5y ! 7 !4y 2 + 3y 2 2 2 2 105. 5 ( 0.3x ! 0.7y ) ! 3( 0.6x ! 0.9y ) 107. 7 ( 0.25a ! 0.46b ) ! 6 (1.14a ! 2.15b ) 1 % 1 % "1 "2 109. 4 $ x ! y' ! 6 $ x ! y' #3 #3 6 & 3 & 1"2 4 % 1" 3 3 % 111. $ x ! y' ! $ x ! y' 2# 3 5 & 3# 4 8 & 293 ( ) ( ) 102. 3( x ! 5x ! 2 ) ! 4 ( !3x + 2x ! 5 ) 104. !6 ( 3x ! 4x + 5 ) + 4 ( 5x ! 4x ! 6 ) 100. !5 12y 2 ! 8y ! 6 !6y 2 + 13y 2 2 2 2 106. 8 ( 0.7x ! 0.9y ) ! 5 ( 0.3x ! 0.9y ) 108. 6 ( 2.45a ! 4.6b ) ! 9 ( 3.5a ! 5.27b ) 1 % 1 % "1 "1 110. 5 $ x ! y' ! 8 $ x ! y' #2 #3 4 & 5 & 2" 4 3 % 1" 5 1 % 112. $ x ! y' ! $ x ! y' 3# 5 4 & 3# 6 2 &