Step-by-step Solution:

To determine the number of values of \( k \) for which the given system of equations has infinitely many solutions, let's analyze the equations: \[ (k+1)x + 8y = 4k \] \[ kx + (k+3)y = 3k - 1 \] Step 1: Condition for Infinite Solutions
For a system of two linear equations to have infinitely many solutions, the two equations must be proportional, meaning their coefficients must be in the same ratio: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] where the general form of a linear equation is: \[ a_1x + b_1y = c_1 \] \[ a_2x + b_2y = c_2 \] From the given system:
\( a_1 = k+1 \), \( b_1 = 8 \), \( c_1 = 4k \)
\( a_2 = k \), \( b_2 = k+3 \), \( c_2 = 3k - 1 \)
Thus, the condition for infinitely many solutions is: \[ \frac{k+1}{k} = \frac{8}{k+3} = \frac{4k}{3k - 1} \] Step 2: Solve for \( k \)
Setting the first two ratios equal: \[ \frac{k+1}{k} = \frac{8}{k+3} \] Cross multiplying: \[ (k+1)(k+3) = 8k \] \[ k^2 + 3k + k + 3 = 8k \] \[ k^2 + 4k + 3 = 8k \] \[ k^2 - 4k + 3 = 0 \] Factoring: \[ (k-3)(k-1) = 0 \] \[ k = 3 \quad \text{or} \quad k = 1 \] Now, check with the third ratio: \[ \frac{8}{k+3} = \frac{4k}{3k - 1} \] For \( k = 3 \): \[ \frac{8}{6} = \frac{12}{8} \quad \text{(not equal)} \] For \( k = 1 \): \[ \frac{8}{4} = \frac{4}{2} = 2 \] which is valid. So, only \( k = 1 \) satisfies the condition.
Final Answer:
The number of values of \( k \) for which the system has infinitely many solutions is 1, which matches option B (1).