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DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A LESSON 2 . 3 Name Class Date Solving Absolute Value 2.3 Solving Absolute Value Inequalities Inequalities Essential Question: What are two ways to solve an absolute value inequality? Resource Common Core Math Standards Locker The student is expected to: Explore COMMON A-CED.A.1 Visualizing the Solution Set of an Absolute CORE Value Inequality Create equations and inequalities in one variable and use them to solve You know that when solving an absolute value equation, it’s possible to get two solutions. problems. Also A-REI.B.3, F-IF.C.7b Here, you will explore what happens when you solve absolute value inequalities. Mathematical Practices A Determine whether each of the integers from -5 to 5 is a solution COMMON Number Solution? of the inequality x + 2 < 5. Write yes or no for each number CORE MP.4 Modeling ⎜ ⎟ x = -5 no in the table. If a number is a solution, plot it on the number line. x = -4 no Language Objective x = -3 no Match absolute value equations and inequalities with their graphs, x = -2 yes explaining and justifying reasoning. - - - - - 5 4 3 2 1 0 1 2 3 4 5 x = -1 yes x = 0 yes x = 1 yes ENGAGE x = 2 yes x = 3 no Essential Question: What are two x = 4 no ways to solve an absolute value x = 5 no inequality? y an Determine whether each of the integers from -5 to 5 is a solution B Number Solution? Possible answer: You can solve an absolute value mp of the inequality x + 2 > 5. Write yes or no for each number o ⎜ ⎟ x = -5 yes C g in the table. If a number is a solution, plot it on the number line. n inequality graphically or algebraically. For a i h x = -4 yes s i l graphical solution, treat each side of the inequality ub x = -3 no P t ur o x = -2 no as a function and graph the two functions. Use the c ar H - - - - - x = -1 n 5 4 3 2 1 0 1 2 3 4 5 no i l inequality symbol to determine the intervals on the f f i x = 0 no M n o x-axis where one graph lies above or below the t x = 1 no gh u other. For an algebraic solution, isolate the absolute o x = 2 no H © x = 3 value expression and rewrite the inequality as a no compound inequality that doesn’t involve absolute x = 4 yes value so that you can finish solving the inequality. x = 5 yes " o Module 2 ile inf 87 Lesson 3 ough "F PREVIEW: LESSON thr --Changes must be made T EDIT A DO NO A;CA- - ectionKey=NL orr C e t Da Class PERFORMANCE TASK Name alue e V bsolut Solving A e 2 . 3 c Resour er ock Inequalities alue inequality? L e v . e an absolut oblems HARDCHARDCOOVERVER PAGES 6370 o solv s t e pr y a o w o solv e tw t ar ha W uestion: ariable and use them t tial Q Essen e tions and inequalities in one v bsolut e equa t ea OMMON .A.1 Cr C A-CED et of an A ORE C .C.7b View the Engage section online. Discuss the names of .3, F-IF olution S Also A-REI.B isualizing the S V y e s. n tio u l Explor o so et tw o g e t l alue Inequalit b i V s s pos ’ t s. n, i tie tio ali ua u q eq ue e e in l u e va l t e va u t A2_MNLESE385894_U1M02L3.indd 87 l u 19/03/14 1:49 PM so l o b s n a b e a v g a l vin o l u s olution? n so o S e n y h e t w s wh a n h pe Number w t p o a n t h u k a o h n Y e w o r ti no o u l l u will exp s a so -5 o o 5 i = , y 5 t x e r - e H m o s fr r r be e eg no t um e in h n h c t a f r e -4 h o o c f = the planets, the elliptical path the planets follow, and r ea no x e or . eth yes e h e t r lin e w ri be in . W um no rm 5 Turn to these pages to e < e n Det + 2 n th -3 x⎟ t o = ⎜ t i x o ty l uali n, p o eq ti u l e in a so s f th s ye o r i be 2 um = - f a n x . I e l b e ta in th = -1 yes x yes 4 5 x = 0 s 1 0 1 2 3 = 1 ye 2- x find this lesson in the 3- 4 - s 5- ye why the distance a planet is from the sun might relate - x = 2 = 3 no x no x = 4 = 5 no x hardcover student to absolute value inequalities. Then preview the Number Solution? n s o ye ti u l s a so = -5 o 5 i x m -5 t o s fr r r be s e eg ye t um n h n e i c f th r ea -4 h o o = c f x r ea no e or . eth yes e edition. h e lin t r e w ri be no in . W rm 5 um e > e n y Det ⎟ + 2 n th = -3 x t o x ⎜ t i o ty l an uali n, p o eq ti u l e in o no f th s a s o r i mp be -2 um x = o f a n . I e l b e ta Cgn in th = -1 no ihs x no il = 0 Lesson Performance Task. ub P 2 3 4 5 x no tur 1 0 1 = 1 2 - x o 3- 4 - c 5- no ar H - x = 2 nilf 3 no fi x = s Mn = 4 ye ot x s ghu = 5 ye o H x © n 3 so s Le 87 e 2 19/03/14 1:52 PM ul od M 2L3.indd 87 A2_MNLESE385894_U1M0 87 Lesson 2 . 3 DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through "File info" DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-B;CA-B CorrectionKey=NL-B;CA-B State the solutions of the equation x + 2 = 5 and relate them to the solutions you found ⎜ ⎟ for the inequalities in Steps A and B. EXPLORE The solutions are -3 and 3. These are the only numbers that are not solutions of the ⎜ ⎟ ⎜ ⎟ inequalities x + 2 < 5 and x + 2 > 5. Visualizing the Solution Set of an Absolute Value Inequality If x is any real number and not just an integer, graph the solutions of x + 2 < 5 and ⎜ ⎟ x + 2 > 5 . ⎜ ⎟ Graph of all real solutions of x + 2 < 5: Graph of all real solutions of x + 2 > 5: ⎜ ⎟ ⎜ ⎟ QUESTIONING STRATEGIES - - - - - - - - - - How would you characterize the solutions to 5 4 3 2 1 0 1 2 3 4 5 5 4 3 2 1 0 1 2 3 4 5 an absolute value inequality? Sample answer: They lie either between two values or everywhere Reflect else except between the two values, depending on 1. It’s possible to describe the solutions of x + 2 < 5 and x + 2 > 5 using inequalities that don’t ⎜ ⎟ ⎜ ⎟ the inequality. involve absolute value. For instance, you can write the solutions of x + 2 < 5 as x > -3 and ⎜ ⎟ x < 3. Notice that the word and is used because x must be both greater than -3 and less than 3. Why do you write the solutions to the absolute How would you write the solutions of x + 2 > 5? Explain. ⎜ ⎟ ⎜ ⎟ Write the solutions of x + 2 > 5 as x < -3 or x > 3. Use the word or because x must be value inequality as a compound inequality either less than -3 or greater than 3; it can’t be both. statement? Sample answer: because the solution 2. Describe the solutions of x + 2 ≤ 5 and x + 2 ≥ 5 using inequalities that don’t involve consists of the union or intersection of the solutions ⎜ ⎟ ⎜ ⎟ absolute value. of the two related linear inequalities. ⎜ ⎟ The solutions of x + 2 ≤ 5 are the values of x for which x ≥ -3 and x ≤ 3. The solutions ⎜ ⎟ of x + 2 ≥ 5 are the values of x for which x ≤ -3 or x ≥ 3. Explain 1 Solving Absolute Value Inequalities Graphically EXPLAIN 1 ( ) ( ) ( ) ( ) You can use a graph to solve an absolute value inequality of the form ƒ x > g x or ƒ x < g x , ( ) ( ) where ƒ x is an absolute value function and g x is a constant function. Graph each function © Solving Absolute Value Inequalities H separately on the same coordinate plane and determine the intervals on the x-axis where one o u ( ) ( ) gh Graphically graph lies above or below the other. For ƒ x > g x , you want to find the x-values for which t o n ( ) ( ) ( ) ( ) M the graph ƒ x is above the graph of g x . For ƒ x < g x , you want to find the x-values for i f f ( ) ( ) l which the graph of ƒ x is below the graph of g x . i n H ar c QUESTIONING STRATEGIES o Example 1 Solve the inequality graphically. y ur t P ub How do you interpret which points are l i s x + 3 + 1 > 4 h ⎜ ⎟ 2 i n g solutions of the inequality? Sample answer: C x o ( ) ( ) mp The inequality is of the form ƒ x > g x , so determine the intervals on If f(x) > g(x), the graph of f(x) must be “above” the - - - 0 an ( ) 6 4 2 the x-axis where the graph of ƒ x = x + 3 + 1 lies above the graph of ⎜ ⎟ ( ) y g x = 4. -2 graph of g(x). The solutions are the x-values in the ( ) ( ) interval along the x-axis where f(x) has y-values The graph of ƒ x = x + 3 + 1 lies above the graph of g x = 4 ⎜ ⎟ -4 to the left of x = -6 and to the right of x = 0, so the solution of greater than g(x). If f(x) < g(x), the graph of f(x) x + 3 + 1 > 4 is x < -6 or x > 0. ⎜ ⎟ must be “below” the graph of g(x). The solutions are the x-values in the interval along the x-axis where Module 2 88 Lesson 3 f(x) has y-values less than g(x). PROFESSIONAL DEVELOPMENT How do you know when the endpoints of A2_MNLESE385894_U1M02L3 88 16/05/14 4:41 AM the solution interval on the x-axis are not Math Background included in the solution? The original inequality An absolute value inequality is often in the form |ax + b| < c or |ax + b| > c. is < or >. If the inequality is in the form |ax + b| < c, then it can be rewritten as the compound inequality –c < ax + b < c, and solved for x. The -c - b c - b _ _ solution will be of the form a < x and x < a . If the inequality is in the form |ax + b| > c, it can be rewritten as the compound inequality c < ax + b or c - b _ ax + b < –c, and solved for x. The solution will be of the form a < x or -c - b x <_ . The solutions are easily adjusted for ≤ and ≥. a Solving Absolute Value Inequalities 88 DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-A;CA-A x - 2 - 3 < 1 y B ⎜ ⎟ 4 ( ) ( ) The inequality is of the form f x < g x , so determine the intervals 2 ( ) below x on the x-axis where the graph of f x = x - 2 - 3 lies ( ) ⎜ ⎟ 0 the graph of g x = 1. - 2 2 4 6 ( ) below -2 The graph of f x = x - 2 - 3 lies the graph of ⎜ ⎟ ( ) -4 g x = 1 between x = -2 and x = 6 , so the solution of ⎜ ⎟ x - 2 - 3 < 1 is x > -2 and x < 6 . Reflect 3. Suppose the inequality in Part A is x + 3 + 1 ≥ 4 instead of x + 3 + 1 > 4. How does the ⎜ ⎟ ⎜ ⎟ solution change? The solution now includes the endpoints of the interval: x ≤ -6 or x ≥ 0. 4. In Part B, what is another way to write the solution x > -2 and x < 6? -2 < x < 6 ( ) 5. Discussion Suppose the graph of an absolute value function ƒ x lies entirely above the graph of ( ) ( ) ( ) the constant function g x . What is the solution of the inequality ƒ x > g x ? What is the solution ( ) ( ) of the inequality ƒ x < g x ? ( ) ( ) ( ) The solution of f x > g x is all real numbers, because every point on the graph of f x is ( ) ( ) ( ) above the corresponding point on the graph of g x . The solution of f x < g x is no real ( ) number, because no point on the graph of f x is below the corresponding point on the ( ) graph of g x . Your Turn 6. Solve x + 1 - 4 ≤ -2 graphically. ⎜ ⎟ y ( ) ( ) an The inequality is of the form ƒ x ≤ g x , so determine the intervals on the x-axis mp ( ) ( ) where the graph of ƒ x = x + 1 - 4 intersects or lies below the graph of g x = -2. o ⎜ ⎟ C g n i ( ) ( ) The graph of ƒ x = x + 1 - 4 intersects the graph of g x = -2 at x = -3 and x = 1 h ⎜ ⎟ s i l ( ) ub and lies below the graph of g x = -2 between those x-values, so the solution of P t x + 1 -4 ≤ -2 is x ≥ -3 and x ≤ 1. ur ⎜ ⎟ o c ar H n i l f f i M n o t gh u o H © Module 2 89 Lesson 3 COLLABORATIVE LEARNING A2_MNLESE385894_U1M02L3.indd 89 19/03/14 1:49 PM Small Group Activity Have students work in groups to make a flowchart that explains how to solve each of the four types of an absolute value inequality of the form |ax + b| c or |ax – b| c, where the box represents the inequality symbol. For example: |2x + 3| ≤ 6 |2x + 3| < 6 |2x + 3| > 6 |2x + 3| ≥ 6 Ask each student in a group to finish one branch of the flowchart. Then have the group collate the branches to make the entire flowchart. 89 Lesson 2 . 3 DO NOT EDIT--Changes must be made through “File info” CorrectionKey=NL-C;CA-C DO NOT EDIT--Changes must be made through "File info" DO NOT EDIT--Changes must be made through "File info" CorrectionKey=NL-B;CA-B CorrectionKey=NL-B;CA-B Explain 2 Solving Absolute Value Inequalities Algebraically To solve an absolute value inequality algebraically, start by isolating the absolute value expression. EXPLAIN 2 When the absolute value expression is by itself on one side of the inequality, apply one of the following rules to finish solving the inequality for the variable. Solving Absolute Value Inequalities Solving Absolute Value Inequalities Algebraically Algebraically 1. If x > a where a is a positive number, then x < -a or x > a. ⎜ ⎟ 2. If x < a where a is a positive number, then -a < x < a. ⎜ ⎟ Example 2 Solve the inequality algebraically. Graph the solution on a number line. QUESTIONING STRATEGIES ⎜ ⎟ ⎜ ⎟ When does the graph of the solution to an 4 - x + 15 > 21 x + 4 - 10 ≤ -2 inequality include the endpoints? when the ⎜ ⎟ 4 - x > 6 ⎜ ⎟ x + 4 ≤ 8 original inequality is ≤ or ≥ 4 - x < -6 or 4 - x > 6 x + 4 ≥ -8 and x + 4 ≤ 8 When does the graph of the solution include - x < -10 or - x > 2 x ≥ -12 and x ≤ 4 the points between the endpoints found in the x > 10 or x < -2 The solution is x ≥ -12 and x ≤ 4 , solution? When the original inequality is < or ≤, The solution is x > 10 or x < -2. or ≤ x ≤ 4 . the compound inequality is an “and” statement, so -12 - - - its graph includes the intersection of two graphs. 6 4 2 0 2 4 6 8 10 12 14 - - - 12 8 4 0 4 These graphs intersect between the endpoints. Reflect INTEGRATE TECHNOLOGY ⎜ ⎟ ⎜ ⎟ 7. In Part A, suppose the inequality were 4 - x + 15 > 14 instead of 4 - x + 15 > 21. How would the solution change? Explain. A graphing calculator can be used to check the ⎜ ⎟ The first step in solving would be to subtract 15 from both sides and get 4 - x > -1. solution graph for an inequality. For example, At this point, the solving process can stop, because the absolute value of every number is you would graph y = |4 – x| + 15 > 21 to check the © H greater than -1. So, the solution is all real numbers. o u solution for |4 – x| + 15 > 21. The graph will be a gh t o n M broken horizontal line above the x-axis with i f f l i n 8. In Part B, suppose the inequality were x + 4 - 10 ≤ -11 instead of x + 4 - 10 ≤ -2. H ⎜ ⎟ ⎜ ⎟ endpoints at –2 and 10. You must interpret the graph ar How would the solution change? Explain. c o ur to be open at the endpoints. So, the solution graph is The first step in solving would be to add 10 to both sides and get x + 4 ≤ -1. At this t ⎜ ⎟ P ub l x < –2 or x > 10. i s point, the solving process can stop, because there are no real numbers whose absolute h i n g C value is less than or equal to -1. So, there is no solution. o mp an y AVOID COMMON ERRORS Some students confuse when to use or and and when rewriting an absolute value inequality. Remind students that when the inequality is |ax + b| < c or |ax + b| ≤ c, they should rewrite the inequality using Module 2 90 Lesson 3 and. When the inequality is |ax + b| > c or DIFFERENTIATE INSTRUCTION |ax + b| ≥ c, they should rewrite the inequality using or. Emphasize the importance of checking some of A2_MNLESE385894_U1M02L3.indd 90 15/05/14 5:17 PM the solutions in the original inequality to help avoid Visual Cues You may want students to use visual models to help them understand some simple this error. inequalities as well as some real-world inequalities. For simple inequalities of the form |x| < a or |x|> a, constructing a graph of possible solutions on a number line as a first step may be helpful. For a real-world problem, involving allowable tolerance, for example, drawing an unlabeled conjunction graph centered at the middle value with the endpoints representing the minimum and maximum tolerance values may be most helpful. This should help students visualize how to construct an inequality based on the graph. Solving Absolute Value Inequalities 90
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