- Solve each individual inequality by isolating the variable.
- Combine the results using the 'or' operator, which represents the union of the two sets.
- For 13, 14, 16, interpret the result as a combined set on the number line, often simplifying to a single interval or all real numbers.
- Final results:
- \(\mathbb{R}\)
- \(\mathbb{R}\)
- \((-\infty, 0] \cup (5, \infty)\)
- \(z > -5\)
- \((-\infty, -7] \cup (7, \infty)\)
Explanation
- Understanding Compound Inequalities with 'or'
For each compound inequality involving the logical operator 'or', we find the solution set for each component separately and then take their union. The 'or' operator implies that any value satisfying at least one of the inequalities belongs to the solution set.
- Solving Problem 13
- \(x
eq 6\) or \(x
eq 3\). The inequality \(x
eq 6\) represents all real numbers except 6, while \(x
eq 3\) represents all real numbers except 3. Their union is the set of all real numbers because any real number is either not 6 or not 3 (or both). Thus, the solution is all real numbers, denoted as \(\mathbb{R}\) or \((-\infty, \infty)\).
- Solving Problem 14
- \(2a + 5 > 17\) or \(a + 2 \leq 8\).
- Solve the first: \(2a + 5 > 17 \implies 2a > 12 \implies a > 6\).
- Solve the second: \(a + 2 \leq 8 \implies a \leq 6\).
The union of \(a > 6\) and \(a \leq 6\) covers all possible real values for \(a\). Thus, the solution is all real numbers, \(\mathbb{R}\).
- Solving Problem 15
- \(y - 6 \leq -6\) or \(3y - 5 > 10\).
- Solve the first: \(y - 6 \leq -6 \implies y \leq 0\).
- Solve the second: \(3y - 5 > 10 \implies 3y > 15 \implies y > 5\).
The solution set is the union of these intervals: \((-\infty, 0] \cup (5, \infty)\).
- Solving Problem 16
- \(3z + 5 > 2z\) or \(6z - 3z + 4 \geq 16\).
- Solve the first: \(3z + 5 > 2z \implies z > -5\).
- Solve the second: \(6z - 3z + 4 \geq 16 \implies 3z + 4 \geq 16 \implies 3z \geq 12 \implies z \geq 4\).
The union of \(z > -5\) and \(z \geq 4\) is simply \(z > -5\), because all values \(\geq 4\) are already included in the set \(z > -5\). Thus, the solution is \(z > -5\).
- Solving Problem 18
- \(r - 7 > 0\) or \(2r + 16 \leq 2\).
- Solve the first: \(r - 7 > 0 \implies r > 7\).
- Solve the second: \(2r + 16 \leq 2 \implies 2r \leq -14 \implies r \leq -7\).
The solution set is the union of these two disjoint intervals: \((-\infty, -7] \cup (7, \infty)\).
- Final Results
The solutions are:
- \(x \in \mathbb{R}\)
- \(a \in \mathbb{R}\)
- \(y \in (-\infty, 0] \cup (5, \infty)\)
- \(z > -5\)
- \(r \in (-\infty, -7] \cup (7, \infty)\)
Examples
Compound inequalities are used in logistics for range management. For example, if a delivery truck can only operate efficiently when the temperature is outside the 'danger zone' of 0 to 5 degrees Celsius (i.e., \(\text{Temp} \leq 0\) or \(\text{Temp} > 5\)), it is solving a compound inequality to determine safe operating conditions. Similarly, if a process controller must trigger a warning when a sensor reading falls outside its calibrated safe bounds, it uses these logical structures to maintain system reliability.