49. For what values(s) of \( k \) will the following system of linear equations have no solution? Infinitely many solutions? \[ \left\{\begin{array}{r}2 x-3 y=4 \\ -6 x+9 y=k\end{array}\right. \]

For the system of equations to have no solution, the lines represented by the equations must be parallel, meaning their slopes are the same but their intercepts are different. The first equation can be rearranged to find its slope: \( y = \frac{2}{3}x - \frac{4}{3} \), which has a slope of \( \frac{2}{3} \). The second equation, rewritten in slope-intercept form, will also yield a slope of \( \frac{2}{3} \) when comparing coefficients. For the lines to be parallel, we set the constant term unequal to create the condition \( k \neq 12 \). For the system to have infinitely many solutions, the two equations must represent the same line. This can happen when the second equation can be obtained from the first by multiplying through by a nonzero constant. Here, that would mean \( k = 12 \) to satisfy the equation \( -6 \left(\frac{4}{2}\right) + 9 \left(\frac{4}{3}\right) = k \). Thus, the respective values of \( k \) are \( k \neq 12 \) for no solutions and \( k = 12 \) for infinitely many solutions.