Find the value(s) of p and q for the following pair of equations:
`2x + 3y= 7 and 2px + py= 28-qy`, if the pair of equations have infinitely many solutions.

Answer

To find the values of \( p \) and \( q \) for the equations \( 2x + 3y = 7 \) and \( 2px + py = 28 - qy \) such that they have infinitely many solutions, we will follow these steps:

### Step 1: Write the equations in standard form
The first equation is already in standard form:
\[ 2x + 3y - 7 = 0 \]
From this, we can identify:
- \( A_1 = 2 \)
- \( B_1 = 3 \)
- \( C_1 = -7 \)

For the second equation, rearranging gives:
\[ 2px + py + qy - 28 = 0 \]
This can be rewritten as:
\[ 2px + (p + q)y - 28 = 0 \]
From this, we identify:
- \( A_2 = 2p \)
- \( B_2 = p + q \)
- \( C_2 = -28 \)

### Step 2: Set up the condition for infinitely many solutions
For the equations to have infinitely many solutions, the ratios of the coefficients must be equal:
\[ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} \]
Substituting the values we found:
\[ \frac{2}{2p} = \frac{3}{p + q} = \frac{-7}{-28} \]

### Step 3: Simplify the ratios
From \( \frac{-7}{-28} \), we simplify to:
\[ \frac{7}{28} = \frac{1}{4} \]
Thus, we have:
\[ \frac{2}{2p} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 2 \cdot 4 = 2p \implies 8 = 2p \implies p = 4 \]

### Step 4: Substitute \( p \) back to find \( q \)
Now we substitute \( p = 4 \) into the second ratio:
\[ \frac{3}{p + q} = \frac{1}{4} \]
Substituting \( p = 4 \):
\[ \frac{3}{4 + q} = \frac{1}{4} \]
Cross-multiplying gives:
\[ 3 \cdot 4 = 1 \cdot (4 + q) \implies 12 = 4 + q \]
Solving for \( q \):
\[ q = 12 - 4 = 8 \]

### Final Answer
The values are:
\[ p = 4, \quad q = 8 \]