9.4: Inequalities and Absolute Value ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× × ÷ × ÷ 9.4: Inequalities and Absolute Value Pilar Alcazar Period 1 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |3x+2/4|≤ 5 Since there is a fraction with a denominator of 4, you need to multiply both sides of the equation by 4 or 4/1. Also, the absolute value means you need to do a positive and negative version of the equation. 3x+2/4≤ 5 | 3x+2/4 ≥ -5 x4/1 x4 | x4 x4 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |3x+2/4|≤ 5 2. Now, you need to subtract 2 from both sides of the equation because there is a 2 added onto the 3x. Add 2 to both sides of the second equation because the 2 is negative. 3x+2≤ 20 | 3x+2≥ -20 -2 -2 | -2 -2 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |3x+2/4|≤ 5 3. Now you want to get the x all by itself. You need to divide both sides by 3 to isolate x. In the second equation, divide both sides by negative three and flip the sign from ≥ to ≤. 3x≤ 18 | 3x≥ -22 ÷3 ÷3 | ÷3 ÷3 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |3x+2/4|≤ 5 4. Since x is now isolated, you are finished with the equation. x ≤ 6 | x ≥ -22/3 Answer: {x|-22/3≤x≤6} ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Critical Thinking: Write an absolute value inequality to describe each of the graphs below. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Since the graph only goes to -4 and 2, the inequality would be {x|-4≤x≤2}. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Since the graph is less than -2 or greater than 3, the inequality would be {x|x<-2 or x>3}. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |2b-4|< -5 1. Since this is an absolute value equation, you need to write it in a positive and negative form. 2b-4< -5 | 2b-4> 5 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |2b-4|< -5 2. Now you need to isolate the number that is multiplied onto b and b itself by either adding or subtracting. 2b-4< -5 | 2b-4> 5 +4 +4 | +4 +4 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |2b-4|< -5 3. Now you need to isolate b by dividing by the number that is multiplied onto it. 2b<-1 | 2b>9 ÷2 ÷2 | ÷2 ÷2 ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷× Equation: |2b-4|< -5 Now that b is by itself, you have your answer. b<-1/2 | b>9/2 Answer: {b|b<-1/2 or b>9/2} If you go back and plug the answer in, you find out that these solutions do not validate. Therefore, there is no solution. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

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