Exercise 30 Page 219

The inequality states that the absolute value of 5d+11 is greater than or equal to 3. Since any negative argument inside the absolute value changes signs to be positive, we get a compound inequality when removing the absolute value. Because the distance needs to be greater than or equal to 3, we get an or compound inequality. 5d+11≥ 3 or 5d+11≤- 3 Let's split the compound inequality into two inequalities. First Inequality:& 5d+11≥ 3 Second Inequality:& 5d+11≤- 3We can solve them one at a time.

First Inequality

We can solve inequalities as if they were equations by using inverse operations to isolate the variable.

Although d≥- 85 is a perfectly valid expression, we can also rewrite the fraction as a mixed number.

The solution set to this inequality contains all the values greater than or equal to - 1 35. {d | d≥- 1 35}

Second Inequality

We can solve the second inequality in the same way.

Although d≤- 145 is a perfectly valid expression, we can also rewrite the fraction as a mixed number.

The solution set to this inequality contains all the values less than or equal to - 2 45. {d | d≤- 2 45}

The Union of Sets

The solution to the compound inequality joined by the word or is the union of the solution sets from the first and second inequalities. {d | d≥- 1 35} ⋃ {d | d≤- 2 45}