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9.4: Inequalities and Absolute Value Pilar Alcazar Period 1

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Equation: |3x+2/4|≤ 5 1.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|x4x4

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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≥ |-2 -2

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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

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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}

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

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

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

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

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

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

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Equation: |2b-4|< -5 4.Now that b is by itself, you have your answer. b 9/2 Answer: {b|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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