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Try to rewrite this inequality as a compound inequality.
Solution Set: -6 ≤ v ≤ 3
Graph:
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 18 away from the midpoint in the positive direction and any number less than or equal to 18 away from the midpoint in the negative direction. Abs. Value Inequality:& |4v+6|≤ 18 Compound Inequality:& - 18≤ 4v+6 ≤ 18 This compound inequality means that the distance from 4v+6 is greater than or equal to - 18 and less than or equal to 18. 4v+6≥- 18 and 4v+6≤ 18 Let's isolate v in both of these cases before graphing the solution set.
Rearrange inequality
LHS-6≤RHS-6
.LHS /4.≤.RHS /4.
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: & v ≤ 3 Second Solution Set: -6 ≤ & v Intersecting Solution Set: -6 ≤ & v ≤ 3
The graph of this inequality includes all values from -6 to 3, inclusive. We show this by using closed circles on the endpoints.