We can split the compound inequality into two cases, connecting them with an AND.
15≤ 2x+5 AND 2x+5 ≤ 17
We will solve the two cases one at the time and then recombine them. Let's start with the first simple inequality.
This inequality says that all values less than or equal to 6 will satisfy the inequality. 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:& 5 ≤ x
Second Solution Set:& x ≤ 6
Intersecting Solution Set:& 5 ≤ x ≤ 6
Let's now graph the inequality to get a better idea of the solutions. Since the symbols are less than or equal to there will be closed circles on 5 and 6. Since x is greater than 5, the graph will be shaded to the right of 5. Since x is less than 6, the graph will be shaded to the left of 6.
The solutions are all numbers that are between 5 and 6, inclusive.
Error
Because we cannot see the student's calculations or graph, we cannot be certain of the error which was made. However, if the student solved correctly algebraically, then they could have misinterpreted the meaning of the resulting compound inequality. For example, they could have thought it was an OR inequality.
5≤ x OR x≤ 6 *
The solution for the above compound inequality is all real numbers.
