If `log_(y)x+log_(x)y=2,x^(2)+y=12`, then the values of `x,y` are

Options

A. `(3,3)`
B. `(-4,-4)`
C. `(4,8)`
D. `(1,11)`

Answer

To solve the problem, we need to find the values of \(x\) and \(y\) given the equations:

1. \( \log_y x + \log_x y = 2 \)
2. \( x^2 + y = 12 \)

### Step 1: Rewrite the logarithmic equation

Using the property of logarithms, we can rewrite \( \log_y x \) as \( \frac{1}{\log_x y} \). Therefore, we have:

\[ \log_y x + \log_x y = 2 \implies \log_y x + \frac{1}{\log_y x} = 2 \]

Let \( t = \log_y x \). Then the equation becomes:

\[ t + \frac{1}{t} = 2 \]

### Step 2: Multiply through by \( t \)

To eliminate the fraction, multiply both sides by \( t \):

\[ t^2 + 1 = 2t \]

### Step 3: Rearrange the equation

Rearranging gives us a standard quadratic equation:

\[ t^2 - 2t + 1 = 0 \]

### Step 4: Factor the quadratic

This can be factored as:

\[ (t - 1)^2 = 0 \]

### Step 5: Solve for \( t \)

Setting the factor equal to zero gives:

\[ t - 1 = 0 \implies t = 1 \]

### Step 6: Substitute back to find \( x \) and \( y \)

Since \( t = \log_y x = 1 \), we can write:

\[ \log_y x = 1 \implies x = y^1 \implies x = y \]

### Step 7: Substitute into the second equation

Now substitute \( x = y \) into the second equation:

\[ x^2 + y = 12 \implies x^2 + x = 12 \]

### Step 8: Rearrange the equation

This can be rearranged to:

\[ x^2 + x - 12 = 0 \]

### Step 9: Factor the quadratic

Factoring gives:

\[ (x - 3)(x + 4) = 0 \]

### Step 10: Solve for \( x \)

Setting each factor to zero gives:

\[ x - 3 = 0 \implies x = 3 \]
\[ x + 4 = 0 \implies x = -4 \]

### Step 11: Determine valid solutions

Since \( x \) and \( y \) must be greater than 0 (as logarithms are only defined for positive values), we discard \( x = -4 \). Thus:

\[ x = 3 \implies y = 3 \]

### Final Answer

The values of \( x \) and \( y \) are:

\[ \boxed{(3, 3)} \]