Which of the following is a solution to the equation `y=sqrt(15-y)+3`?
I.-1
II. 6
III. 11

Options

A. II only
B. III only
C. I and II only
D. II and III only

Answer

To solve the equation \( y = \sqrt{15 - y} + 3 \) and determine which of the given options is a solution, we will follow these steps:

### Step 1: Rearranging the Equation
We start by isolating the square root term:
\[ y - 3 = \sqrt{15 - y} \]

### Step 2: Squaring Both Sides
Next, we square both sides to eliminate the square root:
\[ (y - 3)^2 = 15 - y \]

### Step 3: Expanding the Left Side
Now, we expand the left side:
\[ y^2 - 6y + 9 = 15 - y \]

### Step 4: Rearranging the Equation
We move all terms to one side of the equation:
\[ y^2 - 6y + 9 + y - 15 = 0 \]
This simplifies to:
\[ y^2 - 5y - 6 = 0 \]

### Step 5: Factoring the Quadratic Equation
Now we will factor the quadratic equation:
\[ (y - 6)(y + 1) = 0 \]

### Step 6: Finding the Solutions
Setting each factor to zero gives us the possible solutions:
\[ y - 6 = 0 \quad \Rightarrow \quad y = 6 \]
\[ y + 1 = 0 \quad \Rightarrow \quad y = -1 \]

### Step 7: Checking the Solutions
We need to check if these solutions satisfy the original equation \( y = \sqrt{15 - y} + 3 \).

1. **For \( y = 6 \)**:
\[ 6 = \sqrt{15 - 6} + 3 \quad \Rightarrow \quad 6 = \sqrt{9} + 3 \quad \Rightarrow \quad 6 = 3 + 3 \quad \Rightarrow \quad 6 = 6 \quad \text{(True)} \]

2. **For \( y = -1 \)**:
\[ -1 = \sqrt{15 - (-1)} + 3 \quad \Rightarrow \quad -1 = \sqrt{16} + 3 \quad \Rightarrow \quad -1 = 4 + 3 \quad \Rightarrow \quad -1 = 7 \quad \Rightarrow \quad \text{(False)} \]

### Conclusion
The only valid solution to the equation is \( y = 6 \). Therefore, the answer is:
- **Option II: 6**