Consider the following equation in x and y: `(x-2y-1)^2 + (4x+3y-4)^2 + (x-2y-1)(4x+3y-4) =0` How many solutions to `(x, y)` with `x, y` real, does the equation have

Options

A. zero
B. exactly one
C. exactly two
D. more than two

Explanation

To solve the equation \((x-2y-1)^2 + (4x+3y-4)^2 + (x-2y-1)(4x+3y-4) = 0\) and determine how many real solutions \((x, y)\) exist, we can follow these steps:

### Step 1: Analyze the equation
The equation is a sum of squares and a product of two terms. Since squares are always non-negative, the only way for the entire expression to equal zero is if each individual term is zero.

### Step 2: Set each term to zero
We can set:
1. \( (x - 2y - 1)^2 = 0 \)
2. \( (4x + 3y - 4)^2 = 0 \)
3. \( (x - 2y - 1)(4x + 3y - 4) = 0 \)

From the first two equations, we can derive:
- From \( (x - 2y - 1)^2 = 0 \):
\[ x - 2y - 1 = 0 \implies x = 2y + 1 \]

- From \( (4x + 3y - 4)^2 = 0 \):
\[ 4x + 3y - 4 = 0 \implies 4x = 4 - 3y \implies x = 1 - \frac{3}{4}y \]

### Step 3: Equate the two expressions for \(x\)
Now we have two expressions for \(x\):
1. \( x = 2y + 1 \)
2. \( x = 1 - \frac{3}{4}y \)

Setting these equal gives:
\[ 2y + 1 = 1 - \frac{3}{4}y \]

### Step 4: Solve for \(y\)
Rearranging the equation:
\[ 2y + \frac{3}{4}y = 1 - 1 \]
\[ \frac{8y}{4} + \frac{3y}{4} = 0 \]
\[ \frac{11y}{4} = 0 \implies y = 0 \]

### Step 5: Substitute \(y\) back to find \(x\)
Substituting \(y = 0\) back into either equation for \(x\):
\[ x = 2(0) + 1 = 1 \]

### Step 6: Check the solution against the original conditions
We have found one potential solution: \((x, y) = (1, 0)\). However, we must ensure that the denominator condition from the identity we used is satisfied:
\[ 3x + 5y - 3 \neq 0 \]
Substituting \(x = 1\) and \(y = 0\):
\[ 3(1) + 5(0) - 3 = 3 - 3 = 0 \]
This means that the solution \((1, 0)\) is not valid since it makes the denominator zero.

### Conclusion
Since we derived a potential solution that does not satisfy the conditions of the original equation, we conclude that there are no valid solutions for \((x, y)\) in real numbers.

Thus, the final answer is:
\[ \text{Number of real solutions: } 0 \]