For what value of p the pair of linear equations `(p + 2)x – (2p + 1)y = 3(2p – 1)` and `2x – 3y = 7` has a unique solution.

Answer

To determine the value of \( p \) for which the pair of linear equations

\[ (p + 2)x - (2p + 1)y = 3(2p - 1) \]
and
\[ 2x - 3y = 7 \]

has a unique solution, we will follow these steps:

### Step 1: Write the equations in standard form
We need to express both equations in the standard form \( Ax + By + C = 0 \).

1. For the first equation:
\[ (p + 2)x - (2p + 1)y - 3(2p - 1) = 0 \]
Simplifying this gives:
\[ (p + 2)x - (2p + 1)y - (6p - 3) = 0 \]
Rearranging, we have:
\[ (p + 2)x - (2p + 1)y + (3 - 6p) = 0 \]

2. For the second equation:
\[ 2x - 3y - 7 = 0 \]

### Step 2: Identify coefficients
From the equations, we identify:
- For the first equation:
- \( A_1 = p + 2 \)
- \( B_1 = -(2p + 1) \)
- \( C_1 = 3 - 6p \)

- For the second equation:
- \( A_2 = 2 \)
- \( B_2 = -3 \)
- \( C_2 = -7 \)

### Step 3: Use the condition for a unique solution
For the system of equations to have a unique solution, the following condition must hold:
\[ \frac{A_1}{A_2} \neq \frac{B_1}{B_2} \]

Substituting the values we found:
\[ \frac{p + 2}{2} \neq \frac{-(2p + 1)}{-3} \]
This simplifies to:
\[ \frac{p + 2}{2} \neq \frac{2p + 1}{3} \]

### Step 4: Cross-multiply to eliminate fractions
Cross-multiplying gives:
\[ 3(p + 2) \neq 2(2p + 1) \]
Expanding both sides:
\[ 3p + 6 \neq 4p + 2 \]

### Step 5: Rearranging the inequality
Rearranging the inequality:
\[ 3p + 6 - 2 \neq 4p \]
This simplifies to:
\[ 4 \neq p \]
or
\[ p \neq 4 \]

### Conclusion
Thus, the value of \( p \) for which the pair of linear equations has a unique solution is any real number except \( 4 \).