- Solve each linear inequality separately to isolate \(x\).
- The first part, \(x + 15 \leq -\frac{1}{3}\), yields \(x \leq -\frac{46}{3}\).
- The second part, \(x + 15 \geq \frac{1}{3}\), yields \(x \geq -\frac{44}{3}\).
- The union of the solutions is \((-\infty, -\frac{46}{3}] \cup [-\frac{44}{3}, \infty)\).
- Final Answer: \(\boxed{(-\infty, -\frac{46}{3}] \cup [-\frac{44}{3}, \infty)}\)
Explanation
- Problem Analysis
The goal is to solve the compound inequality \(x + 15 \leq -\frac{1}{3} \text{ or } x + 15 \geq \frac{1}{3}\) and represent the solution in interval notation. This is a union of two linear inequalities. To find the solution set, we solve each inequality individually for \(x\) and then take the union of the resulting sets.
- Solving the First Inequality
For the first inequality, \(x + 15 \leq -\frac{1}{3}\), subtract 15 from both sides:
\(x \leq -\frac{1}{3} - 15\)
To subtract, find a common denominator:
\(-\frac{1}{3} - \frac{45}{3} = -\frac{46}{3}\)
So, the solution for the first part is \(x \leq -\frac{46}{3}\). In interval notation, this is \((-\infty, -\frac{46}{3}]\).
- Solving the Second Inequality
For the second inequality, \(x + 15 \geq \frac{1}{3}\), subtract 15 from both sides:
\(x \geq \frac{1}{3} - 15\)
Find a common denominator:
\(\frac{1}{3} - \frac{45}{3} = -\frac{44}{3}\)
So, the solution for the second part is \(x \geq -\frac{44}{3}\). In interval notation, this is \([-\frac{44}{3}, \infty)\).
- Final Solution and Representation
The complete solution is the union of the two sets: \(x \leq -\frac{46}{3} \cup x \geq -\frac{44}{3}\). In interval notation, this is expressed as \((-\infty, -\frac{46}{3}] \cup [-\frac{44}{3}, \infty)\). On a number line, this represents two rays: one starting at \(-\frac{46}{3}\) and extending to the left, and another starting at \(-\frac{44}{3}\) and extending to the right, leaving a gap between these two values.
Examples
In chemistry or engineering, compound inequalities often describe the tolerance range for a system's stability. For instance, if a mechanical component requires a pressure level that must either be very low (below a safety threshold) or very high (above an operational threshold) to prevent certain unwanted resonant frequencies, you would solve a compound inequality to determine the valid operational ranges. This ensures that the component remains within safe and efficient bounds.