1. **State the problem:** We want to find the conditions on $a$ and $b$ for the system of linear equations: $$\begin{cases} 2x - 3y = a \\ 4x - 6y = b \end{cases}$$ 2. **Recall the theory:** For a system of two linear equations: $$\begin{cases} A_1x + B_1y = C_1 \\ A_2x + B_2y = C_2 \end{cases}$$ - The system has **no solutions** if the lines are parallel but not coincident, i.e., $$\frac{A_1}{A_2} = \frac{B_1}{B_2} \neq \frac{C_1}{C_2}$$ - The system has **one unique solution** if the lines intersect, i.e., $$\frac{A_1}{A_2} \neq \frac{B_1}{B_2}$$ - The system has **infinitely many solutions** if the two equations represent the same line, i.e., $$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$$ 3. **Identify coefficients:** $$A_1 = 2, \quad B_1 = -3, \quad C_1 = a$$ $$A_2 = 4, \quad B_2 = -6, \quad C_2 = b$$ 4. **Calculate ratios:** $$\frac{A_1}{A_2} = \frac{2}{4} = \frac{1}{2}$$ $$\frac{B_1}{B_2} = \frac{-3}{-6} = \frac{1}{2}$$ Since $$\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{1}{2}$$, the lines are either parallel or coincident. 5. **Check the ratio of constants:** $$\frac{C_1}{C_2} = \frac{a}{b}$$ - If $$\frac{a}{b} \neq \frac{1}{2}$$, then the system has **no solutions** (parallel lines). - If $$\frac{a}{b} = \frac{1}{2}$$, then the system has **infinitely many solutions** (same line). 6. **Summary:** - **No solutions:** $$b \neq 2a$$ - **Infinitely many solutions:** $$b = 2a$$ - **One solution:** This case does not occur here because the ratios of coefficients are equal, so lines are either parallel or coincident. **Final answer:** - The system has no solutions if $$b \neq 2a$$. - The system has infinitely many solutions if $$b = 2a$$. - The system never has exactly one solution for any values of $a$ and $b$ in this case.