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(D) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \

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$p = -1, q = 2$

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(D) For a pair of linear equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Given equations are $4x + 5y = 2$ and $(2p + 7q)x + (p + 8q)y = 2q - p + 1$.Here,$\frac{a_1}{a_2} = \frac{4}{2p + 7q}$,$\frac{b_1}{b_2} = \frac{5}{p + 8q}$,and $\frac{c_1}{c_2} = \frac{2}{2q - p + 1}$.Equating $\frac{a_1}{a_2} = \frac{b_1}{b_2}$:$\frac{4}{2p + 7q} = \frac{5}{p + 8q} \implies 4(p + 8q) = 5(2p + 7q) \implies 4p + 32q = 10p + 35q \implies 6p + 3q = 0 \implies q = -2p$ (Equation $1$).Equating $\frac{a_1}{a_2} = \frac{c_1}{c_2}$:$\frac{4}{2p + 7q} = \frac{2}{2q - p + 1} \implies 4(2q - p + 1) = 2(2p + 7q) \implies 8q - 4p + 4 = 4p + 14q \implies 8p + 6q = 4 \implies 4p + 3q = 2$ (Equation $2$).Substituting $q = -2p$ into Equation $2$:$4p + 3(-2p) = 2 \implies 4p - 6p = 2 \implies -2p = 2 \implies p = -1$.Now,$q = -2(-1) = 2$.Thus,for $p = -1$ and $q = 2$,the equations have infinitely many solutions.

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

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For which values 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$