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 30 Page 46

Try to rewrite this inequality as a compound inequality.

Solution Set: -11/15 ≤ t ≤ 17/15
Graph:

Practice makes perfect
Before we can solve this inequality, we need to isolate the absolute value expression using the Properties of Inequality.
3|5t-1|+9 ≤ 23
3|5t-1| ≤ 14
|5t-1| ≤14/3
We are asked to find and graph the solution set for all possible values of t in the given inequality. |5t-1| ≤ 14/3

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 143 away from the midpoint in the positive direction and any number less than or equal to 143 away from the midpoint in the negative direction. Absolute Value Inequality:& |5t-1| ≤ 14/3 Compound Inequality:& - 14/3≤ 5t-1 ≤ 14/3 We can split this compound inequality into two cases, one where 5t-1 is greater than or equal to - 143 and one where 5t-1 is less than or equal to 143. - 14/3≤5t-1 and 5t-1≤ 14/3 Let's isolate t in both of these cases before graphing the solution set.

Case 1

5t-1≤14/3
Solve for t
5t≤14/3+1
5t≤14/3+3/3
5t≤17/3
t≤.17/3 /5.
t ≤17/15
This inequality tells us that all values less than or equal to 1715 will satisfy the inequality.

Case 2

- 14/3≤ 5t-1
Solve for t
-14/3+1≤ 5t
-14/3+3/3≤ 5t
-11/3≤ 5t
.-11/3 /5.≤ t
-11/15 ≤ t
t ≥ -11/15
This inequality tells us that all values greater than or equal to - 1115 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:& t ≤ 17/15 Second Solution Set:& -11/15 ≤ t Intersecting Solution Set:& -11/15 ≤ t ≤ 17/15

Graph

The graph of this inequality includes all values from - 1115 to 1715, inclusive. We show this by using closed circles on the endpoints.