$$ \frac{a}{-2}

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$$ \frac{a}{-2}<-1 $$ or $$ -4 a+3 \geq 23 $$ Solution Interval Notation $$ \square $$

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Let's solve the compound inequality step by step and express the solution in interval notation. **Given Inequalities:** $$ \frac{a}{-2} < -1 \quad \text{or} \quad -4a + 3 \geq 23 $$ ### 1. Solve the first inequality: $$\frac{a}{-2} < -1$$ $$ \frac{a}{-2} < -1 $$ **Step 1:** Multiply both sides by $$-2$$. **Note:** Multiplying or dividing both sides of an inequality by a negative number reverses the inequality sign. $$ a > 2 \quad \text{(Inequality sign reverses)} $$ ### 2. Solve the second inequality: $$-4a + 3 \geq 23$$ $$ -4a + 3 \geq 23 $$ **Step 1:** Subtract 3 from both sides. $$ -4a \geq 20 $$ **Step 2:** Divide both sides by $$-4$$. **Note:** Dividing by a negative number reverses the inequality sign. $$ a \leq -5 \quad \text{(Inequality sign reverses)} $$ ### 3. Combine the Solutions Since the original inequalities are connected by "or," the solution includes all values that satisfy **either** inequality. $$ a \leq -5 \quad \text{or} \quad a > 2 $$ ### 4. Express in Interval Notation $$ (-\infty, -5] \cup (2, \infty) $$ **Final Answer:** $$ \boxed{\,(-\infty,\ -5\,]\,\cup\, (2,\ \infty)\,} $$ The solution is $$ a \leq -5 $$ or $$ a > 2 $$, which in interval notation is $$ (-\infty, -5] \cup (2, \infty) $$.

\( \frac{a}{-2}<-1 \) or \( -4 a+3 \geq 23 \) Solution Interval Notation \( \square \)

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