Solve the Inequality: 0 < 8a + 4 ≤ -a + 9

To solve this problem, we'll break it down into manageable steps:

The problem asks us to find $a$ satisfying two conditions simultaneously: $0 < 8a + 4$ and $8a + 4 \leq -a + 9$.

• Step 1: Solve the first inequality.

The inequality $0 < 8a + 4$ can be simplified by subtracting 4 from both sides:

$-4 < 8a$

Next, divide each side by 8 to isolate $a$:

$-\frac{1}{2} < a$
• Step 2: Solve the second inequality.

The inequality $8a + 4 \leq -a + 9$ can be simplified. Begin by adding $a$ to both sides to gather all $a$-terms on one side:

$9a + 4 \leq 9$

Subtract 4 from both sides:

$9a \leq 5$

Finally, divide each side by 9 to solve for $a$:

$a \leq \frac{5}{9}$
• Step 3: Combine the results of these inequalities.

We now combine the results from step 1 and step 2:

The condition from step 1 is $-\frac{1}{2} < a$.

The condition from step 2 is $a \leq \frac{5}{9}$.

Together, these conditions provide the range:

$-\frac{1}{2} < a \leq \frac{5}{9}$

The solution set is $-\frac{1}{2} < a \leq \frac{5}{9}$.

Therefore, the correct answer choice is: $-\frac{1}{2} < a \leq \frac{5}{9}$.