For unique solution of $2x + 3y = 7$ and $kx + 6y = 14$, $k$ should NOT equal

For unique solution of $2x + 3y = 7$ and $kx + 6y = 14$, $k$ should NOT equal
  1. $4$
  2. $2$
  3. $3$
  4. $5$

Solution

For coincident: $\dfrac{2}{k} = \dfrac{3}{6} = \dfrac{1}{2} \Rightarrow k = 4$. For unique solution, $k \ne 4$.

Asked in: MH-SSC-9

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