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Try to rewrite this inequality as a compound inequality.
Solution Set: - 3 12 ≤ w ≤ 12
Graph:
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 4 away from the midpoint in the positive direction and any number less than or equal to 4 away from the midpoint in the negative direction. Absolute Value Inequality:& |2w+3| ≤ 4 Compound Inequality:& - 4≤ 2w+3 ≤ 4 We can split this compound inequality into two cases, one where 2w+3 is greater than or equal to -4 and one where 2w+3 is less than or equal to 4. - 4≤2w+3 and 2w+3≤ 4 Let's isolate w in both of these cases before graphing the solution set.
LHS-3≤RHS-3
.LHS /2.≤.RHS /2.
Rearrange inequality
Rewrite 7 as 6+1
Write as a sum of fractions
Calculate quotient
Add terms
Remove parentheses
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:& w ≤ 12 Second Solution Set:& - 3 12 ≤ w Intersecting Solution Set:& - 3 12 ≤ w ≤ 12
The graph of this inequality includes all values from - 3 12 to 12, inclusive. We show this by using closed circles on the endpoints.