Solve and graph the solution set of :
`2x - 9 lt 7 and 3x + 9 le 25, "x" inR`

Answer

To solve the inequalities \(2x - 9 < 7\) and \(3x + 9 \leq 25\), we will follow these steps:

### Step 1: Solve the first inequality \(2x - 9 < 7\)

1. **Add 9 to both sides**:
\[ 2x - 9 + 9 < 7 + 9 \]
This simplifies to:
\[ 2x < 16 \]

2. **Divide both sides by 2**:
\[ x < \frac{16}{2} \]
This simplifies to:
\[ x < 8 \]

### Step 2: Solve the second inequality \(3x + 9 \leq 25\)

1. **Subtract 9 from both sides**:
\[ 3x + 9 - 9 \leq 25 - 9 \]
This simplifies to:
\[ 3x \leq 16 \]

2. **Divide both sides by 3**:
\[ x \leq \frac{16}{3} \]
This simplifies to:
\[ x \leq 5.33 \]

### Step 3: Combine the results

From the first inequality, we have:
\[ x < 8 \]
From the second inequality, we have:
\[ x \leq \frac{16}{3} \quad (\text{which is approximately } 5.33) \]

### Step 4: Determine the solution set

The solution set is the intersection of the two inequalities:
\[ x < 8 \quad \text{and} \quad x \leq \frac{16}{3} \]
Since \(5.33\) is less than \(8\), the more restrictive condition is \(x \leq \frac{16}{3}\).

Thus, the solution set is:
\[ x \in (-\infty, \frac{16}{3}] \]

### Step 5: Graph the solution set

On a number line:
- Draw an open circle at \(8\) (since \(x < 8\) is not included).
- Draw a closed circle at \(\frac{16}{3}\) (since \(x \leq \frac{16}{3}\) is included).
- Shade the region to the left of \(\frac{16}{3}\) extending towards negative infinity.

### Final Answer

The solution set is:
\[ x \in (-\infty, \frac{16}{3}] \]