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(B) For a system 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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$1$

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(B) For a system 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: $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$.Applying the condition: $\frac{k + 1}{k} = \frac{8}{k + 3} = \frac{4k}{3k - 1}$.First,solve $\frac{k + 1}{k} = \frac{8}{k + 3}$:$(k + 1)(k + 3) = 8k \Rightarrow k^2 + 4k + 3 = 8k \Rightarrow k^2 - 4k + 3 = 0$.Factoring gives $(k - 1)(k - 3) = 0$,so $k = 1$ or $k = 3$.Next,check these values in the second equality $\frac{8}{k + 3} = \frac{4k}{3k - 1}$:If $k = 1$: $\frac{8}{1 + 3} = \frac{8}{4} = 2$ and $\frac{4(1)}{3(1) - 1} = \frac{4}{2} = 2$. Since $2 = 2$,$k = 1$ is a solution.If $k = 3$: $\frac{8}{3 + 3} = \frac{8}{6} = \frac{4}{3}$ and $\frac{4(3)}{3(3) - 1} = \frac{12}{8} = \frac{3}{2}$. Since $\frac{4}{3} 
eq \frac{3}{2}$,$k = 3$ is not a solution.Thus,there is only $1$ value of $k$ for which the system has infinitely many solutions.

The number of values of $k$ for which the system of equations $(k + 1)x + 8y = 4k$ and $kx + (k + 3)y = 3k - 1$ has infinitely many solutions,is

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