This pdf includes the following topics:-Distributive Property Speeding Things Up a Bit Simplify Alternate solution Applications Examples
1. 3.4. COMBINING LIKE TERMS 197 3.4 Combining Like Terms We begin our discussion with the definition of a term. Term. A term is a single number or variable, or it can be the product of a number (called its coefficient ) and one or more variables (called its variable part ). The terms in an algebraic expression are separated by addition symbols. You Try It! EXAMPLE 1. Identify the terms in the algebraic expression How many terms are in the algebraic expression 3x2 + 5xy + 9y 2 + 12. 3x2 + 2xy − 3y 2 ? For each term, identify its coefficient and variable part. Solution. In tabular form, we list each term of the expression 3x2 + 5xy + 9y 2 + 12, its coefficient, and its variable part. Term Coefficient Variable Part 3x2 3 x2 5xy 5 xy 2 9y 9 y2 12 12 None Answer: 3
You Try It! EXAMPLE 2. Identify the terms in the algebraic expression How many terms are in the algebraic expression a3 − 3a2 b + 3ab2 − b3 . 11 − a2 − 2ab + 3b2 ? For each term, identify its coefficient and variable part. Solution. The first step is to write each difference as a sum, because the terms of an expression are defined above to be those items separated by addition a3 + (−3a2 b) + 3ab2 + (−b3 ) In tabular form, we list each term of the expression a3 +(−3a2 b)+3ab2 +(−b3 ), its coefficient, and its variable part. Version: Fall 2010
2. 198 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Term Coefficient Variable Part a3 1 a3 −3a2 b −3 a2 b 3ab2 3 ab2 −b3 −1 b3 Answer: 4
Like Terms We define what is meant by “like terms” and “unlike terms.” Like and Unlike Terms. The variable parts of two terms determine whether the terms are like terms or unlike terms. Like Terms. Two terms are called like terms if they have identical variable parts, which means that the terms must contain the same variables raised to the same exponential powers. Unlike Terms. Two terms are called unlike terms if their variable parts are different. You Try It! Are −3xy and 11xy like or EXAMPLE 3. Classify each of the following pairs as either like terms or unlike terms? unlike terms: (a) 3x and −7x, (b) 2y and 3y 2 , (c) −3t and 5u, and (d) −4a3 and 3a3 . Solution. Like terms must have identical variable parts. a) 3x and −7x have identical variable parts. They are “like terms.” b) 2y and 3y 2 do not have identical variable parts (the exponents differ). They are “unlike terms.” c) −3t and 5u do not have identical variable parts (different variables). They are “unlike terms.” d) −4a3 and 3a3 have identical variable parts. They are “like terms.” Answer: Like terms
Version: Fall 2010
3. 3.4. COMBINING LIKE TERMS 199 Combining Like Terms When using the distributive property, it makes no difference whether the mul- tiplication is on the left or the right, one still distributes the multiplication times each term in the parentheses. Distributive Property. If a, b, and c are integers, then a(b + c) = ab + ac and (b + c)a = ba + ca. In either case, you distribute a times each term of the sum. “Like terms” can be combined and simplified. The tool used for combining like terms is the distributive property. For example, consider the expression 3y + 7y, composed of two “like terms” with a common variable part. We can use the distributive property and write 3y + 7y = (3 + 7)y. Note that we are using the distributive property in reverse, “factoring out” the common variable part of each term. Checking our work, note that if we redistribute the variable part y times each term in the parentheses, we are returned to the original expression 3y + 7y. You Try It! EXAMPLE 4. Use the distributive property to combine like terms (if pos- Simplify: −8z − 11z sible) in each of the following expressions: (a) −5x2 − 9x2 , (b) −5ab + 7ab, (c) 4y 3 − 7y 2 , and (d) 3xy 2 − 7xy 2 . Solution. If the terms are “like terms,” you can use the distributive property to “factor out” the common variable part. a) Factor out the common variable part x2 . −5x2 − 9x2 = (−5 − 9)x2 Use the distributive property. 2 = −14x Simplify: −5 − 9 = −5 + (−9) = −14. b) Factor out the common variable part ab. −5ab + 7ab = (−5 + 7)ab Use the distributive property. = 2ab Simplify: −5 + 7 = 2. c) The terms in the expression 4y 3 − 7y 2 have different variable parts (the exponents are different). These are “unlike terms” and cannot be combined. Version: Fall 2010
4. 200 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA d) Factor out the common variable part xy 2 . 3xy 2 − 7xy 2 = (3 − 7)xy 2 Use the distributive property. 2 = −4xy Simplify: 3 − 7 = 3 + (−7) = −4. Answer: −19z
Speeding Things Up a Bit Once you’ve written out all the steps for combining like terms, like those shown in Example 4, you can speed things up a bit by following this rule: Combining Like Terms. To combine like terms, simply add their coefficients and keep the common variable part. Thus for example, when presented with the sum of two like terms, such as in 5x + 8x, simply add the coefficients and repeat the common variable part; that is, 5x + 8x = 13x. You Try It! Combine: −3x2 − 4x2 EXAMPLE 5. Combine like terms: (a) −9y − 8y, (b) −3y 5 + 4y 5 , and (c) −3u2 + 2u2 . Solution. a) Add the coefficients and repeat the common variable part. Therefore, −9y − 8y = −17y. b) Add the coefficients and repeat the common variable part. Therefore, −3y 5 + 4y 5 = 1y 5 . However, note that 1y 5 = y 5 . Following the rule that the final answer should use as few symbols as possible, a better answer is −3y 5 + 4y 5 = y 5 . c) Add the coefficients and repeat the common variable part. Therefore, −3u2 + 2u2 = (−1)u2 . However, note that (−1)u2 = −u2 . Following the rule that the final answer should use as few symbols as possible, a better answer is −3u2 + 2u2 = −u2 . Answer: −7x2
Version: Fall 2010
5. 3.4. COMBINING LIKE TERMS 201 A frequently occurring instruction asks the reader to simplify an expression. Simplify. The instruction simplify is a generic term that means “try to write the expression in its most compact form, using the fewest symbols possible.” One way you can accomplish this goal is by combining like terms when they are present. You Try It! EXAMPLE 6. Simplify: 2x + 3y − 5x + 8y. Simplify: −3a + 4b − 7a − 9b Solution. Use the commutative property to reorder terms and the associative and distributive properties to regroup and combine like terms. 2x + 3y − 5x + 8y = (2x − 5x) + (3y + 8y) Reorder and regroup. = −3x + 11y Combine like terms: 2x − 5x = −3x and 3y + 8y = 11y. Alternate solution. Of course, you do not need to show the regrouping step. If you are more comfortable combining like terms in your head, you are free to present your work as follows: 2x + 3y − 5x + 8y = −3x + 11y. Answer: −10a − 5b
You Try It! EXAMPLE 7. Simplify: −2x − 3 − (3x + 4). Simplify: −9a − 4 − (4a − 8) Solution. First, distribute the negative sign. −2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 −(3x + 4) = −3x − 4. Next, use the commutative property to reorder, then the associative property to regroup. Then combine like terms. = (−2x − 3x) + (−3 − 4) Reorder and regroup. = −5x + (−7) Combine like terms: −2x − 3x = −5x. = −5x − 7 Simplify: −5x + (−7) = −5x − 7. Version: Fall 2010
6. 202 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Alternate solution. You may skip the second step if you wish, simply com- bining like terms mentally. That is, it is entirely possible to order your work as follows: −2x − 3 − (3x + 4) = −2x − 3 − 3x − 4 Distribute negative sign. = −5x − 7 Combine like terms. Answer: −13a + 4
You Try It! Simplify: EXAMPLE 8. Simplify: 2(5 − 3x) − 4(x + 3). −2(3a − 4) − 2(5 − a) Solution. Use the distributive property to expand, then use the commutative and associative properties to group the like terms and combine them. 2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Use the distributive property. = (−6x − 4x) + (10 − 12) Group like terms. = −10x − 2 Combine like terms: −6x − 4x = −10x and 10 − 12 = −2. Alternate solution. You may skip the second step if you wish, simply com- bining like terms mentally. That is, it is entirely possible to order your work as follows: 2(5 − 3x) − 4(x + 3) = 10 − 6x − 4x − 12 Distribute. = −10x − 2 Combine like terms. Answer: −4a − 2
You Try It! Simplify: EXAMPLE 9. Simplify: −8(3x2 y − 9xy) − 8(−7x2 y − 8xy). (a2 − 2ab) − 2(3ab + a2 ) Solution. We will proceed a bit quicker with this solution, using the distribu- tive property to expand, then combining like terms mentally. −8(3x2 y − 9xy) − 8(−7x2 y − 8xy) = −24x2 y + 72xy + 56x2 y + 64xy = 32x2 y + 136xy Answer: −a2 − 8ab
Version: Fall 2010
7. 3.4. COMBINING LIKE TERMS 203 We can simplify a number of useful formulas by combining like terms. You Try It! EXAMPLE 10. Find the perimeter P of the (a) rectangle and (b) square A regular hexagon has six pictured below. Simplify your answer as much as possible. equal sides, each with length L s x. Find its perimeter in terms of x. W W s s L s Solution. The perimeter of any polygonal figure is the sum of the lengths of its sides. a) To find the perimeter P of the rectangle, sum its four sides. P = L + W + L + W. Combine like terms. P = 2L + 2W. b) To find the perimeter P of the square, sum its four sides. P = s + s + s + s. Combine like terms. P = 4s. Answer: P = 6x
Sometimes it is useful to replace a variable with an expression containing another variable. You Try It! EXAMPLE 11. The length of a rectangle is three feet longer than twice its The length L of a rectangle width. Find the perimeter P of the rectangle in terms of its width alone. is 5 meters longer than twice Solution. From the previous problem, the perimeter of the rectangle is given its width W . Find the by perimeter P of the rectangle P = 2L + 2W, (3.1) in terms of its width W . where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we’re asked to get the perimeter in terms of the width alone. However, we’re also given the fact that the length is three feet longer than twice the width. Version: Fall 2010
8. 204 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA Three Twice the Length is longer than Feet Width L = 3 + 2W Because L = 3+2W , we can replace L with 3+2W in the perimeter equation 3.1. P = 2L + 2W P = 2(3 + 2W ) + 2W Use the distributive property, then combine like terms. P = 6 + 4W + 2W P = 6 + 6W. Answer: P = 6W + 10 This last equation gives the perimeter P in terms of the width W alone.
You Try It! The width W of a rectangle EXAMPLE 12. The width of a rectangle is two feet less than its length. is 5 feet less than twice its Find the perimeter P of the rectangle in terms of its length alone. width L. Find the perimeter Solution. Again, the perimeter of a rectangle is given by the equation P of the rectangle in terms of its length L. P = 2L + 2W, (3.2) where L and W are the length and width of the rectangle, respectively. This equation gives the perimeter in terms of its length and width, but we’re asked to get the perimeter in terms of the length alone. However, we’re also given the fact that the width is two feet less than the length. Width is Length minus Two feet W = L − 2 Because W = L−2, we can replace W with L−2 in the perimeter equation 3.2. P = 2L + 2W P = 2L + 2(L − 2) Use the distributive property, then combine like terms. P = 2L + 2L − 4 P = 4L − 4. Answer: P = 6L − 10 This last equation gives the perimeter P in terms of the length L alone.
Version: Fall 2010
9. 3.4. COMBINING LIKE TERMS 205 § § § Exercises § § § In Exercises 1-16, combine like terms by first using the distributive property to factor out the common variable part, and then simplifying. 1. 17xy 2 + 18xy 2 + 20xy 2 9. −11x − 13x + 8x 2. 13xy − 3xy + xy 10. −9r − 10r + 3r 3. −8xy 2 − 3xy 2 − 10xy 2 11. −5q + 7q 4. −12xy − 2xy + 10xy 12. 17n + 15n 5. 4xy − 20xy 13. r − 13r − 7r 3 3 6. −7y + 15y 14. 19m + m + 15m 7. 12r − 12r 15. 3x3 − 18x3 8. 16s − 5s 16. 13x2 y + 2x2 y In Exercises 17-32, combine like terms by first rearranging the terms, then using the distributive property to factor out the common variable part, and then simplifying. 17. −8 + 17n + 10 + 8n 25. −14x2 y − 2xy 2 + 8x2 y + 18xy 2 18. 11 + 16s − 14 − 6s 26. −19y 2 + 18y 3 − 5y 2 − 17y 3 19. −2x3 − 19x2 y − 15x2 y + 11x3 27. −14x3 + 16xy + 5x3 + 8xy 20. −9x2 y − 10y 3 − 10y 3 + 17x2 y 28. −16xy + 16y 2 + 7xy + 17y 2 21. −14xy − 2x3 − 2x3 − 4xy 29. 9n + 10 + 7 + 15n 3 3 22. −4x + 12xy + 4xy − 12x 30. −12r + 5 + 17 + 17r 23. −13 + 16m + m + 16 31. 3y + 1 + 6y + 3 24. 9 − 11x − 8x + 15 32. 19p + 6 + 8p + 13 In Exercises 33-56, simplify the expression by first using the distributive property to expand the ex- pression, and then rearranging and combining like terms mentally. 33. −4(9x2 y + 8) + 6(10x2 y − 6) 37. −s + 7 − (−1 − 3s) 3 3 34. −4(−4xy + 5y ) + 6(−5xy − 9y ) 38. 10y − 6 − (−10 − 10y) 35. 3(−4x2 + 10y 2 ) + 10(4y 2 − x2 ) 39. −10q − 10 − (−3q + 5) 36. −7(−7x3 + 6x2 ) − 7(−10x2 − 7x3 ) 40. −2n + 10 − (7n − 1) Version: Fall 2010
10. 206 CHAPTER 3. THE FUNDAMENTALS OF ALGEBRA 41. 7(8y + 7) − 6(8 − 7y) 49. −8(−n + 4) − 10(−4n + 3) 42. −6(−5n − 4) − 9(3 + 4n) 50. 3(8r − 7) − 3(2r − 2) 43. 7(10x2 − 8xy 2 ) − 7(9xy 2 + 9x2 ) 51. −5 − (10p + 5) 44. 10(8x2 y − 10xy 2 ) + 3(8xy 2 + 2x2 y) 52. −1 − (2p − 8) 45. −2(6 + 4n) + 4(−n − 7) 53. 7(1 + 7r) + 2(4 − 5r) 46. −6(−2 − 6m) + 5(−9m + 7) 54. (5 − s) + 10(9 + 5s) 47. 8 − (4 + 8y) 55. −2(−5 − 8x2 ) − 6(6) 48. −1 − (8 + s) 56. 8(10y 2 + 3x3 ) − 5(−7y 2 − 7x3 ) 57. The length L of a rectangle is 2 feet 60. The width W of a rectangle is 9 feet longer than 6 times its width W . Find shorter than its length L. Find the the perimeter of the rectangle in terms of perimeter of the rectangle in terms of its its width alone. length alone. 58. The length L of a rectangle is 7 feet 61. The length L of a rectangle is 9 feet longer than 6 times its width W . Find shorter than 4 times its width W . Find the perimeter of the rectangle in terms of the perimeter of the rectangle in terms of its width alone. its width alone. 59. The width W of a rectangle is 8 feet 62. The length L of a rectangle is 2 feet shorter than its length L. Find the shorter than 6 times its width W . Find perimeter of the rectangle in terms of its the perimeter of the rectangle in terms of length alone. its width alone. § § § Answers § § § 1. 55xy 2 15. −15x3 3. −21xy 2 17. 2 + 25n 5. −16xy 19. 9x3 − 34x2 y 7. 0 21. −18xy − 4x3 9. −16x 23. 3 + 17m 25. −6x2 y + 16xy 2 11. 2q 27. −9x3 + 24xy 13. −19r Version: Fall 2010
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