1. **State the problem:** Find the value of $k$ for which the system of equations $$kx - y = 2$$ and $$6x - 2y = 3$$ 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:** - A system has **no solution** if the lines are parallel but not coincident. - A system has **infinitely many solutions** if the two equations represent the same line. 3. **Rewrite both equations in slope-intercept form $y = mx + b$ to compare slopes and intercepts:** From the first equation: $$kx - y = 2 \implies -y = -kx + 2 \implies y = kx - 2$$ From the second equation: $$6x - 2y = 3 \implies -2y = -6x + 3 \implies y = \frac{6x}{2} - \frac{3}{2} = 3x - \frac{3}{2}$$ 4. **Compare slopes and intercepts:** - First line slope: $m_1 = k$, intercept: $b_1 = -2$ - Second line slope: $m_2 = 3$, intercept: $b_2 = -\frac{3}{2}$ 5. **No solution condition:** Lines are parallel but not the same line, so slopes equal but intercepts different: $$k = 3 \quad \text{and} \quad -2 \neq -\frac{3}{2}$$ So, for no solution: $$\boxed{k = 3}$$ 6. **Infinitely many solutions condition:** Lines are the same, so slopes and intercepts equal: $$k = 3 \quad \text{and} \quad -2 = -\frac{3}{2}$$ Since $-2 \neq -\frac{3}{2}$, 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.