Question

What is the value of $k$ for which the following system of equations has no solution: $2x - 8y = 3$ and $kx + 4y = 10$

• A. $-2$
• B. $1$
• C. $-1$
• D. $2$

Hint:

For parallel lines with no intersection, slopes must match but intercept ratios must differ.

Answer 1

The Correct Option is D

For a system of linear equations: \[ a_1x + b_1y = c_1,\quad a_2x + b_2y = c_2 \] No solution exists if: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \] Here: \[ a_1 = 2,\quad b_1 = -8,\quad c_1 = 3 \] \[ a_2 = k,\quad b_2 = 4,\quad c_2 = 10 \] Condition: \[ \frac{2}{k} = \frac{-8}{4} \quad \Rightarrow \quad \frac{2}{k} = -2 \quad \Rightarrow \quad k = -1 \] But this makes $\frac{a_1}{a_2} = \frac{b_1}{b_2}$ AND $\frac{c_1}{c_2} = \frac{3}{10}$, which is not equal to $-2$, so no solution occurs. Wait, check: No solution requires equal slope ratio but different constant ratio. If $k=2$: \[ \frac{2}{2} = 1,\quad \frac{-8}{4} = -2 \quad \Rightarrow \quad \text{Not parallel},\quad \text{incorrect} \] Re-evaluating — Actually for no solution: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \] \[ \frac{2}{k} = \frac{-8}{4} = -2 \quad \Rightarrow \quad 2/k = -2 \quad \Rightarrow \quad k = -1 \] Then $\frac{c_1}{c_2} = 3/10 \neq -2$, so indeed no solution occurs for $k=-1$. Thus correct option is (c) $-1$. \fbox{Final Answer: (c) $-1$} %Quick tip