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Record and Practice Journal Answer Key 5. − 7 26 6. 1 5 22 Value When 8. − 45.45 7. 9.452 Expression 9 ft 10 x = 0 x = 1 10. −3 9. 7.9 a. 4 4 4 b. −x + 1 1 0 c. 4x − 4 −4 0 d. 2x + 6 6 8 e. 5 x − 13 −13 −8 f. − 2 x + 10 10 8 g. x + 2 2 3 h. 2x − 1 −1 1 i. − 2x + 2 2 0 j. −x + 2 2 1 Chapter 3 Fair Game Review 1. 6 2. 12 3. 20 19 3 4. or 4 4 4 5. 12 6. −1 7. 24 ft2 8. 8 + y 9. p − 6 11. 11c − 8 10. 7m r 12. r − 2 13. 9( z + 4) 3.1 Activity 1. a. Value When Expression 6 A. 3x + 2 − x + 4 B. 5( x − 3) + 2 x = 0 x = 1 6 8 −13 −8 C. x + 3 − ( 2 x + 1) 2 1 D. − 4 x + 2 − x + 3x 2 0 E. − (1 − x ) + 3 2 3 F. 2 x + x − 3x + 4 4 4 G. 4 − 3 + 2( x − 1) −1 1 H. 2(1 − x + 4) 10 8 I. 5 − (4 − x + 2 x) 1 0 J. 5 x − (2 x + 4 − x) −4 0 Big Ideas Math Red Answers A matches d; B matches e; C matches j; D matches i; E matches g; F matches a; G matches h; H matches f; I matches b; J matches c. b. Sample answer: The expressions in the right column have no parentheses and do not have more than one x-term or more than one constant term. To obtain the equivalent expressions, the like terms are combined. 2. Answer should include, but is not limited to: Students should describe the steps they can use to simplify an expression. Then students should use 3 expressions from Activity 1 as examples. Finally, students should write 3 new expressions as exercises. Copyright © Big Ideas Learning, LLC All rights reserved. Record and Practice Journal Answer Key 3. Sample answer: You can simplify an algebraic 4. a. expression by: (1) using the Distributive Property to eliminate parentheses, (2) use the Commutative and Associative Properties of Addition to get the variable terms together and the terms without variables together, (3) combine the variable terms and combine the terms without variables. Sample answer: 3( x − 2) + 5 − x = 3 x − 6 + 5 − x (1) = 3x − x − 6 + 5 (2) = 2x − 1 (3) (2 x b. c. 2. Terms: − 9, 2.5 y, − 0.7 y, 1, 6.4 y 2 Like terms: 2.5 y and − 0.7 y; − 9 and 1 6. 7 d − 5 7. 11g + 8k 5. − 0.7 w + 1.1 8. 4 p − 8 9. 20 w + 9m ) ( ) − 6 ) + (3 x + 2 ) = 5 x − 4 ) ( ) + 4) − ( x + 2) = x + 2 ) ( ) + 3) − ( 2 x − 1) = 2 x + 4 6. ( − x + 3) − ( − 2 x − 2) = x + 5 3.2 Practice 3.2 Activity 1. a. x + 3 + 1) + ( x − 1) = 3 x using algebraic tiles. To add: Combine “like” tiles and remove zero pairs. Write the resulting algebraic expression modeled by the remaining tiles. To subtract: Remove any “like” tiles from the first expression and the second expression. If tiles remain in the second expression, add zero pairs to the first expression as necessary. Then remove any “like” tiles. Write the resulting algebraic expression modeled by the remaining tiles. Like terms: 3x and − 7 x; 4 and − 6 2 3 ) 5. Sample answer: Model the algebraic expressions 1. Terms: 3 x, 4, − 7 x, − 6 4. − 3m + ( ( (4 x 3.1 Practice 3. 3a + 9 ( (2 x d. ) ( (2 x 4. Sample answer: It may be easier to evaluate an algebraic expression for a given value when the expression is simplified. For real-life problems, it may be easier to recognize a value, like a total cost, based on the parts of a simplified algebraic expression. ( b. 2 x − 2 c. 2 x + 3 2. a. (x + 2) + ( x + 4) = 2 x + 6 b. (x − 5) + ( x − 2) = 2 x − 7 c. (x + 5) + ( 2 x − 3) = 3x + 2 d. ( 2 x − 8) + (3 x + 5) = 5 x − 3 3. a. (x + 3) − ( x + 1) = 2 b. (x − 4) − ( x − 3) = −1 c. ( 2 x + 5) − ( x − 1) = x + 6 d. 3 x − 4 1. 2 x + 4 2. − 2n − 1 3. − 4 y + 5 4. −13k − 7 5. 3 g − 4 6. 3r − 2 7. 2 x + 17 3.2 Extension Practice 1. 7(1 + 4) 2. 25(1 + 2) 3. 7(b − 1) 4. 8( a − 2) 5. 4( 2 x + 3) 6. 12( y + 2t ) 7. 10( w + 5 z ) 8. 2(5v + 6u ) 9. 3(3a + 5b) 10. 1 (a − 1) 2 13. 3⎛ 4⎞ ⎜y − ⎟ 10 ⎝ 3⎠ d. (3 x − 7) − ( 2 x + 3) = x − 10 Copyright © Big Ideas Learning, LLC All rights reserved. 11. 1 (d − 3) 4 12. 5⎛ 4⎞ ⎜s + ⎟ 6⎝ 5⎠ 14. 1.1( x + 9) Big Ideas Math Red Answers 7 Record and Practice Journal Answer Key 15. 3.4(c + 3) 16. − 2(3 x − 5) 1⎛ 9⎞ 17. − ⎜ − y + ⎟ 3⎝ 2⎠ 18. ( 2 x + 3) ft 3.3 Activity 1. a. 3.3 Practice 1. −38 2. − 27 4. −13.1 5. 5 3. −37 21 40 6. −8.72 7. x − 42 = − 50; − 8 8. 32 = z + 9; 23 = 10. −198 9. −$1.8 million = 3 ft 4 3.4 Activity 1. a. = = = = = red; 3 yellow; 3; So, x = −1. b. z = 8 2. a. n = − 7 c. b = − 8 c. p = 4 = d. t = −10 = b. y = −15 3; red; 3; x = − 4 d. z = − 4 3. a. − 4 = x + 1; x = −5 c. x − 5 = − 4; x = 1 b. x − 3 = 3; x = 6 d. 5 = x − 2; x = 7 4. a. x + 57 = 19; − 38°F b. Sample answer: The melting point of bromine is 19 degrees Fahrenheit. This is 57 degrees Fahrenheit more than the melting point of mercury. So, subtract 57 from 19 to get 19 − 57 = − 38. So, the melting point of mercury is − 38°F. The method in part (a) is the algebraic solution because it assigns a variable to the unknown quantity. The method in part (b) is the arithmetic solution because it is a direct numerical solution that does not use any variables. 5. Sample answer: Use algebra tiles to model the equation. Then get the variable tile by itself by adding tiles and removing zero pairs. 6. Sample answer: To solve an addition equation, subtract to get the variable by itself. To solve a subtraction equation, add to get the variable by itself. b. k = − 4 c. t = −5 d. m = − 4 e. h = − 4 2. a. 8 = 4 x, x = 2 b. 6 x = −12, x = − 2 c. −10 = 2 x, x = −5 d. 3 x = −18, x = − 6 3. a. 32 x = 400; 12.5 sec; Divide the distance by the rate to get b. c. 400 = 12.5. 32 x = − 96; − 345.6°F; Multiply the melting point 3.6 of radon by 3.6 to get − 96 • 3.6 = − 345.6. 3 x = − 6; − $8 million; Multiply this year’s 4 3 profit by the reciprocal of to get 4 4 −6 • = − 8. 3 Sample answer: The method that uses the equation is the algebraic solution because it assigns a variable to the unknown quantity. The other method is the arithmetic solution because it is a direct numerical solution that does not use any variables. 7. The value of x changes or varies, so x can equal both 2 and 3 in two different problems. 8. Sample answer: The weather varies from day to day. The amount of food a person eats in a day varies. 8 Big Ideas Math Red Answers Copyright © Big Ideas Learning, LLC All rights reserved. Record and Practice Journal Answer Key 4. Sample answer: Multiplication can be used to solve an equation that uses division and division can be used to solve an equation that uses multiplication. The equation x ÷ ( −3) = − 6 uses division, so multiply each side by −3 to obtain x = 18. The equation 2 x = −10 uses multiplication, so divide each side by 2 to obtain x = − 5. 3.4 Practice 1. −30 3. 30 4. − 0.25 5. −50 6. 1.1 x = 7; −56 −8 b. $38.92 5. Add enough 1 or −1 tiles to each side to remove the zero pairs and leave only variable tiles on one side. Form as many equal groups of 1 or −1 tiles as there are variable tiles. One of these groups shows the value of the variable. 6. Subtract 1 from each side. 7. x = 25; The steps are the same. 3 2. − 4 7. 4. a. 25 points 8. −12 x = 60; −5 3.5 Practice 1. −3 2. 6 3. − 7 4. 3 5. − 6.9 6. 1 7. a. A = 2 w − 3 9. 65 cups 2 9 b. 7 meters Chapter 4 3.5 Activity Fair Game Review 1. = 1. 2. = −6 −5 −4 −3 −2 −1 0 −7 −6 −5 −4 −3 −2 −1 −1 0 1 2 3 4 5 4 5 6 7 8 9 10 3. = 4. = 5. = −2.5 −2.4 −2.3 −2.2 −2.1 −2 −1.9 6. = 0 3; 3; 3; 2; 2; x = −1 1 5 2 5 3 5 4 5 1 6 5 7. 413 414 415 416 417 418 419 2. 2x − 3 = −5 Write the equation. 2x − 3 + 3 = −5 + 3 Add 3 to each side. 2x = − 2 Simplify. −2 2x = 2 2 Divide each side by 2. 8. > 9. > 10. > 11. < 12. < 13. < 14. your friend; 5.6 ft is about 5 ft and 7 in. 4.1 Activity 1. a. 5, 6; The numbers are greater than or equal to 5; a. adding 3 to each side yes; The phrase at least implies that the number can be equal to 5; Sample answer: 7, 8, 9, 10 b. − 5, − 6, − 7; The numbers are less than − 4; no; The phrase less than implies that the number cannot be equal to − 4; Sample answer: − 8, − 9, −10, −11 b. They are the same. c. 4, 5, 6; The numbers are greater than 3; no; The x = −1 So, x = −1. 3. a. 2 x + 2 = − 6; x = − 4 b. −13 = 3 x − 4; x = −3 Copyright © Big Ideas Learning, LLC All rights reserved. Simplify. phrase more than implies that the number cannot be equal to 3; Sample answer: 7, 8, 9, 10 Big Ideas Math Red Answers 9