Step-by-step Solution:

These equations are of the form: \[ a_1 x + b_1 y = c_1 \] \[ a_2 x + b_2 y = c_2 \] where:
- \( a_1 = k+1 \), \( b_1 = 8 \), \( c_1 = 4k \)
- \( a_2 = k \), \( b_2 = k+3 \), \( c_2 = 3k - 1 \)
For infinite solutions, the ratios must be equal: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Find \(\frac{a_1}{a_2} = \frac{k+1}{k}\) \[ \frac{k+1}{k} = 1 + \frac{1}{k} \] Find \(\frac{b_1}{b_2} = \frac{8}{k+3}\)
Find \(\frac{c_1}{c_2} = \frac{4k}{3k-1}\)
Setting the first two ratios equal: \[ \frac{k+1}{k} = \frac{8}{k+3} \] Cross multiplying: \[ (k+1)(k+3) = 8k \] \[ k^2 + 3k + k + 3 = 8k \] \[ k^2 - 4k + 3 = 0 \] \[ (k-3)(k-1) = 0 \] So, \( k = 3 \) or \( k = 1 \).
Check if these satisfy \(\frac{a_1}{a_2} = \frac{c_1}{c_2}\)
For \( k = 3 \): \[ \frac{c_1}{c_2} = \frac{4(3)}{3(3)-1} = \frac{12}{9-1} = \frac{12}{8} = \frac{3}{2} \] \[ \frac{a_1}{a_2} = \frac{3+1}{3} = \frac{4}{3} \] Since \( \frac{4}{3} \neq \frac{3}{2} \), \( k=3 \) does not work.
For \( k = 1 \): \[ \frac{c_1}{c_2} = \frac{4(1)}{3(1)-1} = \frac{4}{3-1} = \frac{4}{2} = 2 \] \[ \frac{a_1}{a_2} = \frac{1+1}{1} = \frac{2}{1} = 2 \] Since both ratios are equal, \( k=1 \) satisfies the condition.
There is one valid value of \( k \), which is \( k = 1 \).