For what value of $k$ does the system $kx+3y=k-3$ and $12x+ky=k$ have infinitely many solutions?
Step-by-Step Solution
Key Concept: Set up the equal-ratios condition and solve, then verify the extraneous root.
$\dfrac{a_1}{a_2}=\dfrac{k}{12}$ and $\dfrac{b_1}{b_2}=\dfrac{3}{k}$. Setting these equal: $\dfrac{k}{12}=\dfrac{3}{k}\Rightarrow k^2=36\Rightarrow k=\pm6$. [1.0 Mark]
Check $k=6$: $\dfrac{c_1}{c_2}=\dfrac{6-3}{6}=\dfrac12$, and $\dfrac{a_1}{a_2}=\dfrac{6}{12}=\dfrac12$ — matches. [1.0 Mark]
Check $k=-6$: $\dfrac{c_1}{c_2}=\dfrac{-6-3}{-6}=\dfrac32$, but $\dfrac{a_1}{a_2}=\dfrac{-6}{12}=-\dfrac12$ — does NOT match, so $k=-6$ is rejected. Hence $k=6$. [1.0 Mark]
Correct Answer: