9 4 inequalities and absolute value
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÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×. × ÷ × ÷ × ÷ ×. × ÷ × ÷ × ÷ ×. 9.4: Inequalities and Absolute Value. Pilar Alcazar Period 1. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×. ÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×. Equation: |3x+2/4|≤ 5.

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9 4 inequalities and absolute value
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9.4: Inequalities and Absolute Value

Pilar Alcazar

Period 1

÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×

equation 3x 2 4 5
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 51
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 52
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 53
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×Critical Thinking:Write an absolute value inequality to describe each of the graphs below.

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since the graph is less than 2 or greater than 3 the inequality would be x x 2 or x 3
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 51
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 52
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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 53
÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×÷×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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