For what value of p will the following pair of linear equations have infinitely many solutions.
`(p-3) x+ 3y= p`
`px+ py= 12`

Answer

To find the value of \( p \) for which the given pair of linear equations has infinitely many solutions, we start with the equations:

1. \( (p-3)x + 3y = p \) (Equation 1)
2. \( px + py = 12 \) (Equation 2)

### Step 1: Identify coefficients
We can rewrite the equations in the standard form \( a_1x + b_1y = c_1 \) and \( a_2x + b_2y = c_2 \).

From Equation 1:
- \( a_1 = p - 3 \)
- \( b_1 = 3 \)
- \( c_1 = p \)

From Equation 2:
- \( a_2 = p \)
- \( b_2 = p \)
- \( c_2 = 12 \)

### Step 2: Set up the condition for infinitely many solutions
For the two equations to have infinitely many solutions, the following condition must hold:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

### Step 3: Set up the ratios
Using the coefficients identified:
1. \( \frac{a_1}{a_2} = \frac{p-3}{p} \)
2. \( \frac{b_1}{b_2} = \frac{3}{p} \)
3. \( \frac{c_1}{c_2} = \frac{p}{12} \)

### Step 4: Equate the first two ratios
Set the first two ratios equal:
\[ \frac{p-3}{p} = \frac{3}{p} \]

Cross-multiplying gives:
\[ (p - 3) \cdot p = 3 \cdot p \]
\[ p^2 - 3p = 3p \]
\[ p^2 - 6p = 0 \]

### Step 5: Factor the equation
Factoring gives:
\[ p(p - 6) = 0 \]

### Step 6: Solve for \( p \)
Setting each factor to zero gives:
1. \( p = 0 \)
2. \( p = 6 \)

### Step 7: Verify the second ratio
Now, we need to check the second ratio:
\[ \frac{3}{p} = \frac{p}{12} \]

Cross-multiplying gives:
\[ 3 \cdot 12 = p^2 \]
\[ 36 = p^2 \]
\[ p = 6 \quad \text{or} \quad p = -6 \]

### Conclusion
The values of \( p \) that allow the system to have infinitely many solutions are \( p = 6 \) and \( p = -6 \).