If `(sqrt(3))^(x+y)=9` and `(sqrt(2))^(x-y)=32`, then `2x+y` is _______.

Options

A. 1
B. 0
C. 17
D. 11

Explanation

To solve the equations given in the problem, we will follow these steps:

1. **Rewrite the equations using powers of 2 and 3:**
- The first equation is \((\sqrt{3})^{x+y} = 9\).
- We know that \(9 = 3^2\), and \(\sqrt{3} = 3^{1/2}\).
- Therefore, we can rewrite the equation as:
\[ (3^{1/2})^{x+y} = 3^2 \]
- This simplifies to:
\[ 3^{(1/2)(x+y)} = 3^2 \]
- Since the bases are the same, we can equate the exponents:
\[ \frac{1}{2}(x+y) = 2 \]

2. **Solve for \(x+y\):**
- Multiply both sides by 2:
\[ x+y = 4 \quad \text{(Equation 1)} \]

3. **Now, rewrite the second equation:**
- The second equation is \((\sqrt{2})^{x-y} = 32\).
- We know that \(32 = 2^5\), and \(\sqrt{2} = 2^{1/2}\).
- Therefore, we can rewrite the equation as:
\[ (2^{1/2})^{x-y} = 2^5 \]
- This simplifies to:
\[ 2^{(1/2)(x-y)} = 2^5 \]
- Again, since the bases are the same, we equate the exponents:
\[ \frac{1}{2}(x-y) = 5 \]

4. **Solve for \(x-y\):**
- Multiply both sides by 2:
\[ x-y = 10 \quad \text{(Equation 2)} \]

5. **Now we have a system of equations:**
- From Equation 1: \(x + y = 4\)
- From Equation 2: \(x - y = 10\)

6. **Add the two equations to eliminate \(y\):**
\[ (x+y) + (x-y) = 4 + 10 \]
- This simplifies to:
\[ 2x = 14 \]

7. **Solve for \(x\):**
\[ x = 7 \]

8. **Substitute \(x\) back into Equation 1 to find \(y\):**
\[ 7 + y = 4 \]
- Rearranging gives:
\[ y = 4 - 7 = -3 \]

9. **Now we can find \(2x + y\):**
\[ 2x + y = 2(7) + (-3) = 14 - 3 = 11 \]

Thus, the value of \(2x + y\) is **11**.