Absolute Value Inequalities - CCfaculty.org

1 - Absolute Value InequalitiesObjective: Solve, graph and give interval notation for the solution toinequalities with Absolute an inequality has an Absolute Value we will have to remove the absolutevalue in order to graph the solution or give interval notation. The way we removethe Absolute Value depends on the direction of the inequality |x|< Value is defined as distance from zero. Another way to read thisinequality would be the distance from zero is less than 2. So on a number line wewill shade all points that are less than 2 units away from graph looks just like the graphs of the three part compound Inequalities !When the Absolute Value isless thana number we will remove the Absolute valueby changing the problem to a three part inequality, with the negative Value on theleft and the positive Value on the right. So|x|<2becomes 2< x <2, as thegraph above |x|> Value is defined as distance from zero.

2 Another way to read thisinequality would be the distance from zero is greater than on the numberline we shade all points that are more than 2 units away from graph looks just like the graphs of the OR compound Inequalities ! When theabsolute Value isgreater thana number we will remove the Absolute Value bychanging the problem to an OR inequality, the first inequality looking just likethe problem with no Absolute Value , the second flipping the inequality symbol andchanging the Value to a negative. So|x|>2becomesx >2orx < 2, as the graphabove View Note:The phrase Absolute Value comes from German mathemati-cian Karl Weierstrass in 1876, though he used the Absolute Value symbol for com-plex numbers. The first known use of the symbol for integers comes from a 1939edition of a college algebra text!1 For all Absolute Value Inequalities we can also express our answers in intervalnotation which is done the same way it is done for standard compound can solve Absolute Value Inequalities much like we solvedabsolute Value equa-tions.

3 Our first step will be to isolate the Absolute Value . Next we will remove theabsolute Value by making a three part inequality if the Absolute Value is less thana number, or making an OR inequality if the Absolute Value is greater than anumber. Then we will solve these inequalites. Remember, if we multiply or divideby a negative the inequality symbol will switch directions!Example , graph, and give interval notation for the solution|4x 5|>6 Absolute Value is greater,use OR4x 5>6OR4x 56 6 Solve+ 5 + 5+ 5 + 5 Add5to both sides4x>11 OR4x6 1 Divide both sides by44444x>114 ORx6 14 Graph( , 14] [114, )Interval notationExample , graph, and give interval notation for the solution 4 3|x|6 16 Add4to both sides+ 4+ 4 3|x|6 12 Divide both sides by 3 3 3 Dividing byanegative switches the symbol|x|>4 Absolute Value is greater,use ORx>4 ORx6 4 Graph2( , 4] [4, )Interval NotationIn the previous example, we cannot combine 4and 3because they are not liketerms, the 3has an Absolute Value attached.

4 So we must first clear the 4byadding 4, then divide by 3. The next example is , graph, and give interval notation for the solution9 2|4x+ 1|>3 Subtract9from both sides 9 9 2|4x+ 1|> 6 Divide both sides by 2 2 2 Dividing by negative switches the symbol|4x+ 1|<3 Absolute Value is less,use three part 3<4x+ 1<3 Solve 1 1 1 Subtract1from all three parts 4<4x <2 Divide all three parts by4444 1< x <12 Graph( 1,12)Interval NotationIn the previous example, we cannot distribute the 2into the Absolute can never distribute or combine things outside the Absolute Value with what isinside the Absolute Value . Our only way to solve is to first isolate the absolutevalue by clearing the values around it, then either make a compound inequality(and OR or a three part) to is important to remember as we are solving these equations, the Absolute valueis always positive. If we end up with an Absolute Value is lessthan a negativenumber, then we will have no solution because Absolute valuewill always be posi-3tive, greater than a negative.

5 Similarly, if Absolute valueis greater than a nega-tive, this will always happen. Here the answer will be all real , graph, and give interval notation for the solution12+ 4|6x 1|<4 Subtract 12 from both sides 12 124|6x 1|< 8 Divide both sides by444|6x 1|< 2 Absolute Value can tbe less thananegativeNo Solution or Example , graph, and give interval notation for the solution5 6|x+ 7|617 Subtract5from both sides 5 5 6|x+ 7|612 Divide both sides by 6 6 6 Dividing byanegative flips the symbol|x+ 7|> 2 Absolute Value always greater than negative?All Real Numbers orR?Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) Practice - Absolute Value InequalitiesSolve each inequality, graph its solution, and give interval )|x|<33)|2x|<65)|x 2|<67)|x 7|<39)|3x 2|<911)1 + 2|x 1|6913)6 |2x 5|>= 315)|3x|>517)|x= 3|>= 319)|3x 5|>>321)4 + 3|x 1|>=1023)3 2|x 5|6 1525) 2 3|4 2x|> 827)4 5| 2x 7|< 129)3 2|4x 5|>131) 5 2|3x 6|< 833)4 4| 2x+ 6|> 435)| 10+x|>82)|x|684)|x+ 3|<46)x 8|<128)|x+ 3|6410)|2x+ 5|<912) 10 3|x 2|>414)|x|>516)|x 4|>518)|2x 4|>620)3 |2 x|<122)3 2|3x 1|> 724)4 6| 6 3x|6 526) 3 2|4x 5|>128) 2 + 3|5 x|6430) 2 3| 3x 5> 532)6 3|1 4x|< 334) 3 4| 2x 5|> 7?

6 Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( ) - Absolute Value Inequalities1) 3,32) 8,83) 3,34) 7,15) 4,86) 4,207) 2,48) 7,19) 73,11310) 7,211) 3,512)0,413)1,414)( ,5) (5, )15) ( , 53] [53, )16)( , 1] [9, )17)( , 6) (0, )18) ( , 1) (5, )19)( ,23) (83, )20)( ,0) (4, )21)( , 1] [3, )22)[ 43,2]23)( , 4] [14, )24)( , 52] [ 32, )25)[1,3]26)[12,1]27)( , 4) ( 3, )28)[3,7]29)[1,32]30) [ 2, 43]31)( ,32) (52, )32)( , 12) (1, )33)[2,4]34)[ 3, 2]?Beginning and Intermediate Algebra by Tyler Wallace is licensed under a Creative CommonsAttribution Unported License. ( )6