Exercise 5 Page 494

Try to rewrite this inequality as a compound inequality.

- 36≤ x≤ 28

Practice makes perfect

We are asked to find and graph the solution set for all possible values of x in the given inequality. |x+4|-12≤ 20 First, let's isolate the absolute value expression using the Properties of Inequality.

|x+4|-12≤ 20

AddIneq

LHS+12≤RHS+12

|x+4|≤ 32

Now, 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 32 away from the midpoint in the positive direction and any number less than or equal to 32 away from the midpoint in the negative direction. Absolute Value Inequality:& |x+4|≤ 32 Compound Inequality:& - 32≤ x+4 ≤ 32 We can split this compound inequality into two cases, one where x+4 is greater than or equal to - 32 and one where x+4 is less than or equal to 32. x+4≥- 32 and x+4≤ 32 Let's isolate x in both of these cases before graphing the solution set.

Case 1

x+4≥- 32

SubIneq

LHS-4≥RHS-4

x≥- 36

This inequality tells us that all values greater than or equal to - 36 will satisfy the inequality.

Case 2

x+4≤ 32

SubIneq

LHS-4≤RHS-4

x≤ 28

This inequality tells us that all values less than or equal to 28 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:& - 36≤ x Second Solution Set:& x≤ 28 Intersecting Solution Set:& - 36≤ x≤ 28

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Exercise 5 Page 494

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