Given that the vector `alpha and beta` are non-collinear. The value of x and y for which `u-v=w` holds true if `u= 2 x alpha + y beta, v= 2 y alpha + 3 x beta and w= 2alpha -5 beta`, are

Options

A. `x=2, y=1`
B. `x=1, y=2`
C. `x= -2, y=1`
D. `x= -2, y= -1`

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) such that the equation \( u - v = w \) holds true, where:

- \( u = 2x \alpha + y \beta \)
- \( v = 2y \alpha + 3x \beta \)
- \( w = 2 \alpha - 5 \beta \)

### Step-by-Step Solution:

1. **Set up the equation**:
We start with the equation \( u - v = w \).
\[ (2x \alpha + y \beta) - (2y \alpha + 3x \beta) = 2 \alpha - 5 \beta \]

2. **Simplify the left-hand side**:
Distributing the negative sign:
\[ 2x \alpha + y \beta - 2y \alpha - 3x \beta = 2 \alpha - 5 \beta \]
Combine like terms:
\[ (2x - 2y) \alpha + (y - 3x) \beta = 2 \alpha - 5 \beta \]

3. **Equate coefficients**:
Since \( \alpha \) and \( \beta \) are non-collinear, we can equate the coefficients of \( \alpha \) and \( \beta \) on both sides:
- For \( \alpha \):
\[ 2x - 2y = 2 \quad \text{(Equation 1)} \]
- For \( \beta \):
\[ y - 3x = -5 \quad \text{(Equation 2)} \]

4. **Solve Equation 1**:
From Equation 1:
\[ 2x - 2y = 2 \]
Dividing the entire equation by 2:
\[ x - y = 1 \quad \text{(Equation 3)} \]
Rearranging gives:
\[ y = x - 1 \]

5. **Substitute into Equation 2**:
Substitute \( y = x - 1 \) into Equation 2:
\[ (x - 1) - 3x = -5 \]
Simplifying:
\[ x - 1 - 3x = -5 \]
\[ -2x - 1 = -5 \]
Adding 1 to both sides:
\[ -2x = -4 \]
Dividing by -2:
\[ x = 2 \]

6. **Find \( y \)**:
Substitute \( x = 2 \) back into Equation 3:
\[ y = 2 - 1 = 1 \]

### Final Solution:
Thus, the values of \( x \) and \( y \) are:
\[ x = 2, \quad y = 1 \]