For what positive values of k do the following pair of linear equations have infinitely many solutions?

kx + 3y – ( k – 3 ) = 0

12x + ky – k = 0

The correct option is 6.

To find the positive value of $k$ for which the given system of linear equations has infinitely many solutions, let us analyze the conditions for consistency.

The given pair of linear equations is:

$kx+3y-(k-3)=0$

$12x+ky-k=0$

A general system of two linear equations in two variables:

$a_{1}x+b_{1}y+c_1=0$

$a_{2}x+b_{2}y+c_2=0$

has infinitely many solutions (coincident lines) if and only if the coefficients satisfy the ratio condition:

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

By comparing the coefficients of the given equations with the standard form, we have:

$a_1=k, b_1=3, c_1=-(k-3)$

$a_2=12, b_2=k, c_2=-k$

Substituting these values into the ratio condition gives:

$\frac{k}{12}=\frac{3}{k}=\frac{-(k-3)}{-k}$

First, consider the equality of the first two ratios:

$\frac{k}{12}=\frac{3}{k}$

Cross-multiplying gives:

$k^2=36$

$k=\pm 6$

Next, consider the equality of the second and third ratios:

$\frac{3}{k}=\frac{k-3}{k}$

Since $k\ne 0$, we can multiply both sides by $k$:

$3=k-3$

$k=6$

Thus, the value $k=6$ satisfies all parts of the ratio condition.

Since we are asked for the positive value of $k$, the required answer is 6.