Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
8. Unions and Intersections of Sets
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Exercise 53 Page 220

Write the inequality as a compound inequality and solve each inequality separately.

-5≤ d≤ 5

Practice makes perfect
We are asked to find the solution set for all possible values of d in the given inequality. |4d|≤ 20 To do this, we will create a compound inequality by removing the absolute value. In this case, the distance from the absolute value needs to be less than or equal to 20 in the positive direction and in the negative direction. Absolute Value Inequality:& |4d|≤ 20 Compound Inequality:& - 20≤ 4d≤ 20

We can split this compound inequality into two cases. - 20≤ 4d and 4d≤ 20 Let's isolate d in both of these cases.

Case 1

We can solve inequalities as if they were equations by using inverse operations to isolate the variable.
- 20≤ 4d
- 5≤ d
The solution set to this inequality contains all the values greater than or equal to - 5

Case 2

We can solve the second inequality in the same way.
4d≤ 20
d≤ 5
The solution set to this inequality contains all the values less than or equal to 5.

Solution Set

The solution to the compound inequality is the intersection of the solution sets from the first and second cases. Let's recombine our cases back into one compound inequality. First Solution Set:& - 5≤ d Second Solution Set:& d≤ 5 Intersecting Solution Set:& -5≤ d≤ 5