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Pearson Algebra 2 Common Core, 2011

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Pearson Algebra 2 Common Core, 2011 View details

6. Absolute Value Equations and Inequalities

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Exercise 35 Page 46

Create an or compound inequality because the absolute value is greater than or equal to the given value.

Solution Set: x≤- 3orx≥4
Graph:

Practice makes perfect

Before we can solve this inequality, we need to isolate the absolute value expression using the Division Property of Inequality.

3|2x-1| ≥ 21

.LHS /3.≥.RHS /3.

|2x-1| ≥ 7

We are asked to find and graph the solution set for all possible values of x in the given inequality. |2x-1|≥ 7

To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number greater than or equal to 7 away from the midpoint in the positive direction or any number greater than or equal to 7 away from the midpoint in the negative direction. 2x-1 ≥ 7 or 2x-1 ≤ - 7 Let's isolate x in both of these cases before graphing the solution set.

Case 1

2x-1 ≥ 7

LHS+1≥RHS+1

2x ≥ 8

.LHS /2.≥.RHS /2.

x ≥ 4

This inequality tells us that all values greater than or equal to 4 will satisfy the inequality.

Case 2

2x-1 ≤ - 7

LHS+1≤RHS+1

2x ≤ -6

.LHS /2.≤.RHS /2.

x ≤ -3

This inequality tells us that all values less than or equal to -3 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the combination of the solution sets. First Solution Set:& x ≥4 Second Solution Set:& x ≤ -3 Combined Solution Set:& x ≤ -3 or x ≥ 4

Graph

The graph of this inequality includes all values less than or equal to -3 or greater than or equal to 4. We show this by keeping the endpoints closed.

Subchapter links

Lesson Check p.45

Practice and Problem-Solving Exercises p.46-48

Standardized Test Prep p.48

Mixed Review p.48