For what value of `p and q`, will the following pair of linear equations have infinitely many solutions?
`4x+5y=2`
`(2p+7q)x + (p+8q) y=2q-p+1`

Answer

To find the values of \( p \) and \( q \) for which the given pair of linear equations has infinitely many solutions, we need to apply the condition that the ratios of the coefficients of \( x \), \( y \), and the constant terms must be equal.

The given equations are:
1. \( 4x + 5y = 2 \) (Equation 1)
2. \( (2p + 7q)x + (p + 8q)y = 2q - p + 1 \) (Equation 2)

### Step 1: Identify coefficients
From Equation 1, we can identify:
- \( a_1 = 4 \)
- \( b_1 = 5 \)
- \( c_1 = 2 \)

From Equation 2, we can identify:
- \( a_2 = 2p + 7q \)
- \( b_2 = p + 8q \)
- \( c_2 = 2q - p + 1 \)

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

This gives us three equations to work with.

### Step 3: First ratio
Set up the first ratio:
\[ \frac{4}{2p + 7q} = \frac{5}{p + 8q} \]
Cross-multiplying gives:
\[ 4(p + 8q) = 5(2p + 7q) \]
Expanding both sides:
\[ 4p + 32q = 10p + 35q \]
Rearranging gives:
\[ 10p - 4p + 35q - 32q = 0 \implies 6p + 3q = 0 \]

### Step 4: Second ratio
Now set up the second ratio:
\[ \frac{5}{p + 8q} = \frac{2}{2q - p + 1} \]
Cross-multiplying gives:
\[ 5(2q - p + 1) = 2(p + 8q) \]
Expanding both sides:
\[ 10q - 5p + 5 = 2p + 16q \]
Rearranging gives:
\[ 10q - 16q + 5 + 5p - 2p = 0 \implies 7p - 6q = -5 \]

### Step 5: Solve the system of equations
Now we have a system of equations:
1. \( 6p + 3q = 0 \) (Equation 1)
2. \( 7p - 6q = -5 \) (Equation 2)

From Equation 1, we can express \( q \) in terms of \( p \):
\[ 3q = -6p \implies q = -2p \]

Substituting \( q = -2p \) into Equation 2:
\[ 7p - 6(-2p) = -5 \]
This simplifies to:
\[ 7p + 12p = -5 \implies 19p = -5 \implies p = -\frac{5}{19} \]

Now substituting \( p \) back to find \( q \):
\[ q = -2\left(-\frac{5}{19}\right) = \frac{10}{19} \]

### Final values
Thus, the values of \( p \) and \( q \) for which the equations have infinitely many solutions are:
\[ p = -\frac{5}{19}, \quad q = \frac{10}{19} \]