For what value of $k$ does the system $3x+y=1$ and $(2k-1)x+(k-1)y=2k+1$ have no solution?
Step-by-Step Solution
Key Concept: Apply the no-solution condition: equal first two ratios, unequal third.
For no solution: $\dfrac{3}{2k-1}=\dfrac{1}{k-1}$. Cross-multiplying: $3(k-1)=2k-1\Rightarrow 3k-3=2k-1\Rightarrow k=2$. [1.0 Mark]
Check at $k=2$: $\dfrac{a_1}{a_2}=\dfrac{3}{3}=1$, $\dfrac{c_1}{c_2}=\dfrac{1}{5}$. [1.0 Mark]
Since $\dfrac{a_1}{a_2}=1
eq\dfrac15=\dfrac{c_1}{c_2}$, the no-solution condition is genuinely satisfied at $k=2$. [1.0 Mark]
Correct Answer: