1. **State the problem:** Find the value of $k$ for which the system of equations $$\begin{cases} kx - y = 2 \\ 6x - 2y = 3 \end{cases}$$ has no solution, and determine if there is a value of $k$ for which the system has infinitely many solutions. 2. **Recall the conditions for solutions of linear systems:** - The system has **no solution** if the lines are parallel but not coincident. This means the ratios of coefficients of $x$ and $y$ are equal, but the constants ratio is different. - The system has **infinitely many solutions** if the two equations represent the same line, i.e., all ratios of coefficients and constants are equal. 3. **Write the ratios for the coefficients:** From the system: $$\frac{k}{6} = \frac{-1}{-2} = \frac{2}{3}$$ 4. **Simplify the middle ratio:** $$\frac{-1}{-2} = \frac{1}{2}$$ 5. **Set the ratios for no solution condition:** For no solution, the coefficients ratio must be equal but the constants ratio different: $$\frac{k}{6} = \frac{1}{2} \neq \frac{2}{3}$$ 6. **Solve for $k$:** $$\frac{k}{6} = \frac{1}{2} \implies k = 6 \times \frac{1}{2} = 3$$ 7. **Check constants ratio:** $$\frac{2}{3} \neq \frac{1}{2}$$ So, for $k=3$, the system has no solution. 8. **Check for infinitely many solutions:** All ratios must be equal: $$\frac{k}{6} = \frac{1}{2} = \frac{2}{3}$$ But $$\frac{1}{2} \neq \frac{2}{3}$$ So, there is **no value of $k$** for which the system has infinitely many solutions. **Final answers:** - No solution when $k=3$. - No value of $k$ gives infinitely many solutions.