McGraw Hill Glencoe Algebra 1, 2012
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McGraw Hill Glencoe Algebra 1, 2012 View details
5. Inequalities Involving Absolute Value
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Exercise 23 Page 314

Rewrite this inequality as a compound inequality.

Solution Set: {h | - 5 23 Graph:

Practice makes perfect
We are asked to find and graph the solution set for all possible values of h in the given inequality. | 3h+1/2 |< 8 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 8 away from the midpoint in the positive direction and any number less than 8 away from the midpoint in the negative direction. Absolute Value Inequality:& |3h+1/2|< 8 [0.5em] Compound Inequality:& - 8< 3h+1/2 < 8

We can split this compound inequality into two cases, one where 3h + 12 is greater than -8 and one where 3h + 12 is less than 8. 3h+1/2>- 8 and 3h+1/2< 8 Let's isolate h in both of these cases before graphing the solution set.

Case 1

3h+1/2>- 8
3h+1> - 16
3h> - 17
h>- 17/3
h>-17/3
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Write fraction as a mixed number
h>-15+2/3
h>-(15/3+2/3)
h>-( 5+2/3)
h>-(5 23)
h>- 5 23
This inequality tells us that all values greater than - 5 23 will satisfy the inequality.

Case 2

3h+1/2<8
3h+1<16
3h<15
h<5
This inequality tells us that all values less than 5 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: & - 5 23< h [0.5em] Second Solution Set: & h < 5 [0.5em] Intersecting Solution Set:& - 5 23< h< 5

Graph

The graph of this inequality includes all values from - 5 23 to 5, not inclusive. We show this by using open circles on the endpoints.