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(B) The given pair of linear equations is:$2x + 3y = 7$ ... $(i)$$2px + py = 28 - qy$ ... $(ii)$Rearranging equation $(ii)$,we get:$2px + (p + q)y = 2

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$p = 4, q = 8$

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(B) The given pair of linear equations is:$2x + 3y = 7$ ... $(i)$$2px + py = 28 - qy$ ... $(ii)$Rearranging equation $(ii)$,we get:$2px + (p + q)y = 28$For a pair of linear equations $a_1x + b_1y + c_1 = 0$ and $a_2x + b_2y + c_2 = 0$ to have infinitely many solutions,the condition is $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$.Comparing the given equations with the standard form:$a_1 = 2, b_1 = 3, c_1 = -7$$a_2 = 2p, b_2 = (p + q), c_2 = -28$Applying the condition:$\frac{2}{2p} = \frac{3}{p + q} = \frac{-7}{-28}$$\frac{1}{p} = \frac{3}{p + q} = \frac{1}{4}$Taking the first and third parts:$\frac{1}{p} = \frac{1}{4} \Rightarrow p = 4$Taking the second and third parts:$\frac{3}{p + q} = \frac{1}{4} \Rightarrow p + q = 12$Substituting $p = 4$ into $p + q = 12$:$4 + q = 12 \Rightarrow q = 8$Thus,the values are $p = 4$ and $q = 8$.

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

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