Number of intergeral solution satisfying
`x + sqrt (3-x) ge sqrt(3-x) +3 ` is

Answer

To find the number of integral solutions satisfying the inequality \( x + \sqrt{3 - x} \geq \sqrt{3 - x} + 3 \), we can follow these steps:

### Step 1: Simplify the Inequality
Start by simplifying the given inequality:
\[ x + \sqrt{3 - x} \geq \sqrt{3 - x} + 3 \]
Subtract \( \sqrt{3 - x} \) from both sides:
\[ x \geq 3 \]

### Step 2: Determine the Domain
Next, we need to consider the domain of the square root function. The expression \( \sqrt{3 - x} \) is defined when:
\[ 3 - x \geq 0 \quad \Rightarrow \quad x \leq 3 \]

### Step 3: Combine the Conditions
Now we have two inequalities:
1. \( x \geq 3 \)
2. \( x \leq 3 \)

Combining these gives:
\[ x = 3 \]

### Step 4: Identify Integral Solutions
Since \( x \) must be an integer, the only integral solution is:
\[ x = 3 \]

### Conclusion
Thus, the number of integral solutions satisfying the original inequality is:
\[ \text{Number of solutions} = 1 \]