The number of ordered pairs (x,y) satisfying the system of equations `6^(x)((2)/(3))^(y)-3.2^(x+y)-8.3^(x-y)+24=0,xy=2` is

Options

A. `0`
B. `1`
C. `2`
D. `3`

Answer

To solve the system of equations given by \(6^{x}\left(\frac{2}{3}\right)^{y} - 3 \cdot 2^{x+y} - 8 \cdot 3^{x-y} + 24 = 0\) and \(xy = 2\), we will follow these steps:

### Step 1: Simplify the first equation
We start with the equation:
\[ 6^{x}\left(\frac{2}{3}\right)^{y} - 3 \cdot 2^{x+y} - 8 \cdot 3^{x-y} + 24 = 0 \]
We can express \(6^{x}\) as \( (2 \cdot 3)^{x} = 2^{x} \cdot 3^{x} \) and \(\left(\frac{2}{3}\right)^{y} = \frac{2^{y}}{3^{y}}\). Thus, we rewrite the equation:
\[ 2^{x} \cdot 3^{x} \cdot \frac{2^{y}}{3^{y}} - 3 \cdot 2^{x+y} - 8 \cdot 3^{x-y} + 24 = 0 \]
This simplifies to:
\[ 2^{x+y} \cdot 3^{x-y} - 3 \cdot 2^{x+y} - 8 \cdot 3^{x-y} + 24 = 0 \]

### Step 2: Factor the equation
We can rearrange the equation:
\[ 2^{x+y} \cdot 3^{x-y} - 3 \cdot 2^{x+y} - 8 \cdot 3^{x-y} + 24 = 0 \]
We can factor out \(2^{x+y}\) and \(3^{x-y}\):
\[ 2^{x+y} \cdot (3^{x-y} - 3) - 8 \cdot 3^{x-y} + 24 = 0 \]

### Step 3: Set up equations
We can set:
1. \(2^{x+y} \cdot 3^{x-y} = 8\)
2. \(3^{x-y} - 3 = 0\)

From the second equation, we have:
\[ 3^{x-y} = 3 \implies x-y = 1 \]

### Step 4: Solve for \(x+y\)
From the first equation, we have:
\[ 2^{x+y} \cdot 3 = 8 \implies 2^{x+y} = \frac{8}{3} \]
Taking logarithm base 2:
\[ x+y = \log_{2}\left(\frac{8}{3}\right) = 3 - \log_{2}(3) \]

### Step 5: Solve the system of equations
Now we have the system:
1. \(x - y = 1\)
2. \(x + y = 3 - \log_{2}(3)\)

Adding these two equations:
\[ 2x = 4 - \log_{2}(3) \implies x = 2 - \frac{1}{2}\log_{2}(3) \]
Substituting \(x\) back into \(x - y = 1\):
\[ 2 - \frac{1}{2}\log_{2}(3) - y = 1 \implies y = 1 - \frac{1}{2}\log_{2}(3) \]

### Step 6: Check the second equation \(xy = 2\)
Now we need to check if \(xy = 2\):
\[ xy = \left(2 - \frac{1}{2}\log_{2}(3)\right)\left(1 - \frac{1}{2}\log_{2}(3)\right) \]
Calculating this product and checking if it equals 2 will give us the ordered pairs.

### Step 7: Find all possible pairs
We can also solve the other case \(x + y = 3\) and \(xy = 2\) leading to:
\[ t^2 - 3t + 2 = 0 \implies (t-1)(t-2) = 0 \]
Thus, \(t = 1\) or \(t = 2\) gives us pairs \((1, 2)\) and \((2, 1)\).

### Conclusion
The possible ordered pairs are:
1. \((2, 1)\)
2. \((1, 2)\)
3. \((-1, -2)\)

Thus, the total number of distinct ordered pairs \((x, y)\) that satisfy both equations is \(3\).