Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
7. Absolute Value Equations and Inequalities
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Exercise 43 Page 211

Try to rewrite this inequality as a compound inequality.

Solution Set: -4 ≤ v≤ 5
Graph:

Practice makes perfect
We are asked to find and graph the solution set for all possible values of v in the given inequality. |2v-1|≤9 To do this, we will create a compound inequality by removing the absolute value. In this case, the solution set is any number less than or equal to 9 away from the midpoint in the positive direction and any number less than or equal to 9 away from the midpoint in the negative direction. Absolute Value Inequality:& |2v-1|≤ 9 Compound Inequality:& -9≤ 2v-1 ≤ 9

This compound inequality means that the distance between 2v and 1 is greater than or equal to -9 and less than or equal to 9. 2v-1≥-9 and 2v-1≤9 Let's isolate v in both of these cases before graphing the solution set.

Case 1

2v-1≤9
2v≤10
v≤5
This inequality tells us that all values less than or equal to 5 will satisfy the inequality.

Case 2

2v-1≥-9
2v≥-8
v ≥ -4
This inequality tells us that all values greater than or equal to -4 will satisfy the inequality.

Solution Set

The solution to this type of compound inequality is the overlap of the solution sets. Let's recombine our cases back into one compound inequality. First Solution Set:& v≤ 5 Second Solution Set:& -4≤ v Intersecting Solution Set:& -4≤ v≤ 5

Graph

The graph of this inequality includes all values from -4 to 5, inclusive. We show this by using closed circles on the endpoints.