For what value of $k$ do $kx + 4y = 12$ and $3x + 2y = 6$ have infinitely many solutions?

Explanation

1. 01

   Infinitely many solutions needs $\frac{k}{3} = \frac{4}{2} = \frac{12}{6}$.

2. 02

   $\frac{4}{2} = \frac{12}{6} = 2$, so $\frac{k}{3} = 2$.

3. 03

   Hence $k = 6$. The first equation is then exactly twice the second.