For what value of k, the system of equations $kx + 4y - k + 4 = 0$ and $16x + ky = k$ has an infinite number of solutions?

For a system of two linear equations to have infinite solutions, both equations must represent the same line. This occurs when the coefficients are proportional.

For equations $a_1x + b_1y = c_1$ and $a_2x + b_2y = c_2$, the condition is:

$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$

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Given equations:

$kx + 4y - k + 4 = 0$

$16x + ky = k$

Rewriting the first equation:

$kx + 4y = k - 4$

The second equation is already in standard form:

$16x + ky = k$

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For infinite solutions, the ratios of coefficients must be equal:

$\frac{k}{16} = \frac{4}{k} = \frac{k-4}{k}$

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Using the first two ratios:

$\frac{k}{16} = \frac{4}{k}$

$k \times k = 16 \times 4$

$k^2 = 64$

$k = \pm 8$

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Checking $k = 8$ with the third ratio:

$\frac{4}{k} = \frac{1}{2}$

$\frac{k-4}{k} = \frac{4}{8} = \frac{1}{2}$

The ratios are equal.

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Checking $k = -8$ with the third ratio:

$\frac{4}{k} = \frac{4}{-8} = -\frac{1}{2}$

$\frac{k-4}{k} = \frac{-12}{-8} = \frac{3}{2}$

The ratios are not equal.

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Therefore, $k = 8$.