Lesson 11 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Exit Ticket Sample Solutions 1. Here is the graphical representation of a set of real numbers: a. Describe this set of real numbers in words. The set of all real numbers less than or equal to two b. Describe this set of real numbers in set notation. {π real | π β€ π} (students might use any variable) c. Write an equation or an inequality that has the set above as its solution set. π β π β€ βπ (Answers will, of course, vary, students might use any variable.) 2. Indicate whether each of the following equations is sure to have a solution set of all real numbers. Explain your answers for each. d. π(π + π) = ππ + π No, the two algebraic expressions are not equivalent. e. π + π = π +π Yes, the two expressions are algebraically equivalent by application of the Commutative Property. ππ(π + π) = ππ + πππ f. Yes, the two expressions are algebraically equivalent by application of the Distributive Property. g. ππ(ππ)(ππ) = ππππ No, the two algebraic expressions are not equivalent. Problem Set Sample Solutions For each solution set graphed below, (a) describe the solution set in words, (b) describe the solution set in set notation, and (c) write an equation or an inequality that has the given solution set. 1. 2. a. the set of all real numbers equal to π a. the set of all real numbers equal to c. answers vary c. answers vary b. {π} Lesson 11: Date: © 2013 Common Core, Inc. Some rights reserved. commoncore.org b. π οΏ½ποΏ½ π π Solution Sets for Equations and Inequalities 8/9/13 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 151 Lesson 11 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I 4. 3. a. the set of all real numbers greater than π a. the set of all real numbers less or equal to π c. answers vary c. answers vary b. {π real | π > π} 5. b. {π real | π β€ π} 6. a. the set of all real numbers not equal to π a. the set of all real numbers not equal to π c. answers vary c. answers vary b. {π real | π β π} 7. b. {π real | π β π} 8. a. the null set a. the set of all real numbers b. { } or β b. {π real} c. answers vary Lesson 11: Date: © 2013 Common Core, Inc. Some rights reserved. commoncore.org c. answers vary Solution Sets for Equations and Inequalities 8/9/13 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 152 Lesson 11 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I Fill in the chart below. SOLUTION SET IN WORDS 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. SOLUTION SET IN SET NOTATION The set of real numbers equal to π π = π {π} The set of real numbers equal to π or βπ ππ = π {βπ, π} The set of real numbers not π equal to . ππ β π {π real | z β 1/2} π The set of real numbers equal to π πβ π = π {π} The set of real numbers equal to 1 or -1 ππ + π = π {βπ, π} The set of real numbers equal to 0 π = ππ {π} The set of real numbers greater than 2 π > π {π real | π > 2} The null set πβ π = πβ π { } The set of real numbers less than 4 π β π < βπ {π real | π < 4} The set of real numbers π(π β π) β₯ ππ β π GRAPH β For problems 19β24, answer the following: Are the two expressions algebraically equivalent? If so, state the property (or properties) displayed. If not, state why (the solution set may suffice as a reason) and change the equation, ever so slightly, e.g., touch it up, to create an equation whose solution set is all real numbers. 19. π(π β ππ ) = (βππ + π)π Yes, commutative 20. ππ ππ =π No, the solution set is {π real | π β π}. If we changed it to: numbers. ππ π = ππ, it would have a solution set of all real 21. (π β π)(π + π) + (π β π)(π β π) = (π β π)(ππ β π) Yes, distributive. 22. π π π + = π ππ π No, the solution set is {π}. It we changed it to: Lesson 11: Date: © 2013 Common Core, Inc. Some rights reserved. commoncore.org π π π + = π ππ ππ , it would have a solution set of all real numbers. Solution Sets for Equations and Inequalities 8/9/13 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 153 Lesson 11 NYS COMMON CORE MATHEMATICS CURRICULUM M1 ALGEBRA I 23. ππ + πππ + πππ = πππ No, neither the coefficients nor the exponents are added correctly. One way it could have a solution set of all real numbers would be: ππ + πππ + πππ = ππ (π + ππ + πππ ). 24. π π + ππ + ππ = π(π + π) π Yes, distributive 25. Solve for π: ππ+π π π οΏ½π ππππ | π β π οΏ½ π β π. Describe the solution set in set notation. π 26. Edwina has two sticks, one π yards long and the other π meters long. She is going to use them, with a third stick of some positive length, to make a triangle. She has decided to measure the length of the third stick in units of feet. a. What is the solution set to the statement: βSticks of lengths π yards, π meters and π³ feet make a triangleβ? Describe the solution set in words, and through a graphical representation. One meter is equivalent, to two decimal places, to π. ππ feet. We have that π³ must be a positive length greater than π. ππ feet and less than π. ππ feet. b. What is the solution set to the statement: βSticks of lengths π yards, π meters and π³ feet make an isosceles triangleβ? Describe the solution set in words and through a graphical representation. π³ = π feet or π³ = π. ππ feet. c. What is the solution set to the statement: βSticks of lengths π yards, π meters and π³ feet make an equilateral triangleβ? Describe the solution set in words, and through a graphical representation. The solution set is the empty set. Lesson 11: Date: © 2013 Common Core, Inc. Some rights reserved. commoncore.org Solution Sets for Equations and Inequalities 8/9/13 This work is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. 154
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