- Solve \(6x + 15 \leq 24\) by subtracting 15 then dividing by 6, resulting in \(x \leq 1.5\).
- Solve \(\frac{x}{-2} + 8 \leq 12\) by subtracting 8, then multiplying by -2 (reversing the sign) to get \(x \geq -8\).
- Solve \(\frac{x}{9} - 25 \leq -23\) by adding 25 then multiplying by 9, resulting in \(x \leq 18\).
- Solve \(8x - 2\frac{1}{2} \geq 5\frac{1}{2}\) by adding \(\frac{5}{2}\) to both sides and dividing by 8, giving \(x \geq 1\).
- All solutions are graphed on a number line using a closed circle at the boundary point to represent inclusion.
- Final solutions: \(x \leq 1.5, x \geq -8, x \leq 18, x \geq 1\).
Explanation
- Solving the Inequalities
To solve each linear inequality, we need to isolate the variable \(x\) using algebraic manipulation. Recall the rule that when multiplying or dividing both sides of an inequality by a negative number, the inequality sign must be reversed. We will treat each inequality step-by-step.
- For \(6x + 15 \leq 24\), subtract \(15\) from both sides: \(6x \leq 24 - 15\), which gives \(6x \leq 9\). Dividing by \(6\), we get \(x \leq \frac{9}{6} = 1.5\).
- For \(\frac{x}{-2} + 8 \leq 12\), subtract \(8\) from both sides: \(\frac{x}{-2} \leq 4\). Multiplying both sides by \(-2\) requires us to reverse the inequality sign: \(x \geq 4 \times (-2)\), so \(x \geq -8\).
- For \(\frac{x}{9} - 25 \leq -23\), add \(25\) to both sides: \(\frac{x}{9} \leq 2\). Multiplying by \(9\), we get \(x \leq 18\).
- For \(8x - 2\frac{1}{2} \geq 5\frac{1}{2}\), first convert the mixed numbers to improper fractions: \(2\frac{1}{2} = \frac{5}{2}\) and \(5\frac{1}{2} = \frac{11}{2}\). The inequality becomes \(8x - \frac{5}{2} \geq \frac{11}{2}\). Adding \(\frac{5}{2}\) to both sides: \(8x \geq \frac{11+5}{2} = \frac{16}{2} = 8\). Dividing by \(8\), we get \(x \geq 1\).
- Graphing the Solutions
To graph these solutions on a number line, we represent the solution set visually.
- For \(x \leq 1.5\), we place a closed circle at \(1.5\) and shade to the left.
- For \(x \geq -8\), we place a closed circle at \(-8\) and shade to the right.
- For \(x \leq 18\), we place a closed circle at \(18\) and shade to the left.
- For \(x \geq 1\), we place a closed circle at \(1\) and shade to the right.
A closed circle indicates that the boundary value is included in the solution set, which is appropriate here as all inequalities use \(\leq\) or \(\geq\).
- Summary of Solutions
The solutions for the four inequalities are:
- \(x \leq 1.5\)
- \(x \geq -8\)
- \(x \leq 18\)
- \(x \geq 1\)
Each solution set is represented as a range of values on the number line starting from or ending at the calculated constant.
Examples
Solving linear inequalities is a fundamental skill used in everyday budgeting and resource allocation. For instance, if you have a budget of \(\$100\) to spend on items costing \(\$15\) each with a fixed delivery fee of \(\$10\), you can represent your total cost as \(15x + 10 \leq 100\). Solving this inequality (\(15x \leq 90 \Rightarrow x \leq 6\)) helps you determine the maximum number of items you can purchase without exceeding your budget. Similarly, understanding these ranges helps in planning for time constraints, weight limits on vehicles, or nutritional targets.