A pair of linear equations which has a unique solution x = 2 and y = -3 is

Options

A. `x + y = 1 and 2x - 3y = -5`
B. `2x + 5y = -11 and 4x + 10y= -22`
C. `2x - y =1 and 3x + 2y = 0`
D. `x - 4y - 14 = 0 and 5x - y - 13 = 0`

Explanation

To find a pair of linear equations that has a unique solution \( x = 2 \) and \( y = -3 \), we can start by using the general form of linear equations and substituting the values of \( x \) and \( y \).

### Step 1: Formulate the equations
We can create two linear equations based on the solution \( (2, -3) \). A simple way to do this is to use the point-slope form of a line.

Let’s assume two equations:
1. \( ax + by = c \)
2. \( dx + ey = f \)

We can choose coefficients \( a, b, d, e \) such that when we substitute \( x = 2 \) and \( y = -3 \), we get valid equations.

### Step 2: Create the first equation
Let's create the first equation. We can choose \( a = 1 \), \( b = 1 \), and \( c = -1 \):
\[ 1(2) + 1(-3) = -1 \implies 2 - 3 = -1 \]
Thus, the first equation is:
\[ x + y = -1 \quad \text{(Equation 1)} \]

### Step 3: Create the second equation
Now, let's create the second equation. We can choose \( d = 2 \), \( e = 3 \), and \( f = 0 \):
\[ 2(2) + 3(-3) = 0 \implies 4 - 9 = -5 \]
Thus, the second equation is:
\[ 2x + 3y = -5 \quad \text{(Equation 2)} \]

### Step 4: Verify the solution
Now we have the two equations:
1. \( x + y = -1 \)
2. \( 2x + 3y = -5 \)

We can solve these equations to verify that they yield the solution \( (2, -3) \).

From Equation 1:
\[ y = -1 - x \]

Substituting \( y \) in Equation 2:
\[ 2x + 3(-1 - x) = -5 \]
\[ 2x - 3 - 3x = -5 \]
\[ -x - 3 = -5 \]
\[ -x = -2 \implies x = 2 \]

Now substituting \( x = 2 \) back into Equation 1:
\[ 2 + y = -1 \implies y = -3 \]

### Conclusion
Thus, the pair of linear equations that has a unique solution \( x = 2 \) and \( y = -3 \) is:
1. \( x + y = -1 \)
2. \( 2x + 3y = -5 \)