- Isolate the variable \(x\) in each inequality by performing inverse operations.
- Remember to reverse the inequality sign when multiplying or dividing by a negative number.
- Convert mixed numbers to decimals (e.g., \(2\frac{1}{2} = 2.5\)) to simplify calculations.
- Represent solutions on a number line using a closed circle for \(\leq\) or \(\geq\) and a ray pointing left for \(\leq\) or right for \(\geq\).
- Final results: \(\boxed{x \leq 1.5, x \geq -8, x \leq 18, x \geq 1}\).
Explanation
- Analysis of Inequalities and Rules of Solving
To solve each inequality, we isolate the variable \(x\). The fundamental rules for manipulating inequalities are:
- Adding or subtracting the same value from both sides maintains the direction of the inequality.
- Multiplying or dividing both sides by a positive value maintains the direction of the inequality.
- Multiplying or dividing both sides by a negative value reverses the direction of the inequality.
We will address each of the four provided inequalities individually.
- Solving Inequality 1: \(6x + 15 \leq 24\)
For the first inequality, \(6x + 15 \leq 24\):
- Subtract 15 from both sides: \(6x \leq 24 - 15 = 9\).
- Divide both sides by 6: \(x \leq \frac{9}{6}\).
- Simplifying gives \(x \leq 1.5\).
- Graphing: Place a closed circle at 1.5 and draw a ray pointing to the left, indicating all numbers less than or equal to 1.5.
- Solving Inequality 2: \(\frac{x}{-2} + 8 \leq 12\)
For the second inequality, \(\frac{x}{-2} + 8 \leq 12\):
- Subtract 8 from both sides: \(\frac{x}{-2} \leq 12 - 8 = 4\).
- Multiply both sides by -2: Since we are multiplying by a negative number, the inequality sign flips from \(\leq\) to \(\geq\).
- Thus, \(x \geq 4 \times (-2) = -8\).
- Graphing: Place a closed circle at -8 and draw a ray pointing to the right, indicating all numbers greater than or equal to -8.
- Solving Inequality 6: \(\frac{x}{9} - 25 \leq -23\)
For the sixth inequality, \(\frac{x}{9} - 25 \leq -23\):
- Add 25 to both sides: \(\frac{x}{9} \leq -23 + 25 = 2\).
- Multiply both sides by 9: \(x \leq 2 \times 9 = 18\).
- Graphing: Place a closed circle at 18 and draw a ray pointing to the left, indicating all numbers less than or equal to 18.
- Solving Inequality 7: \(8x - 2\frac{1}{2} \geq 5\frac{1}{2}\)
For the seventh inequality, \(8x - 2\frac{1}{2} \geq 5\frac{1}{2}\):
- Convert the mixed numbers to decimals: \(2\frac{1}{2} = 2.5\) and \(5\frac{1}{2} = 5.5\).
- The inequality is \(8x - 2.5 \geq 5.5\).
- Add 2.5 to both sides: \(8x \geq 5.5 + 2.5 = 8\).
- Divide both sides by 8: \(x \geq \frac{8}{8} = 1\).
- Graphing: Place a closed circle at 1 and draw a ray pointing to the right, indicating all numbers greater than or equal to 1.
- Conclusion
In summary, the solutions to the inequalities are:
- \(x \leq 1.5\)
- \(x \geq -8\)
- \(x \leq 18\)
- \(x \geq 1\)
Each solution is represented graphically on a number line using a closed circle at the boundary point, with a ray pointing in the direction of the inequality. Thus, the final solutions are \(\boxed{x \leq 1.5, x \geq -8, x \leq 18, x \geq 1}\).
Examples
Linear inequalities are essential in real-world budgeting and resource allocation. For example, imagine you are planning a company event with a fixed budget for catering. If the caterer charges a flat setup fee plus a cost per guest, you can use a linear inequality to determine the maximum number of guests you can invite without exceeding your budget. By solving \(Cost \leq Budget\), you ensure the event remains affordable while maximizing your attendance. This mathematical approach helps in making informed, fiscally responsible decisions.