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Mathematics — CBSE Class 10 board question

Q1. [1]
The value of k for which the pair of equations $kx = y + 2$ and $6x = 2y + 3$ has infinitely many solutions,
  1. A is $k = 3$
  2. B does not exist
  3. C is $k = -3$
  4. D is $k = 4$
Previously asked in CBSE board exam
2023 30/1/1 Q2
Generated by claude-sonnet-4-6 · 2026-06-15 10:35 · grounding rag
Model Answer

Answer: B — does not exist

Rewriting: $kx - y - 2 = 0$ and $6x - 2y - 3 = 0$. For infinitely many solutions: $\dfrac{k}{6} = \dfrac{-1}{-2} = \dfrac{-2}{-3}$, i.e., $\dfrac{k}{6} = \dfrac{1}{2}$ gives $k = 3$, but $\dfrac{1}{2} \neq \dfrac{2}{3}$. So no value of $k$ satisfies all three ratios simultaneously; the answer does not exist.

Explanation

For infinitely many solutions, all three ratios must be equal: $\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} = \dfrac{c_1}{c_2}$. Here $\dfrac{b_1}{b_2} = \dfrac{1}{2}$ but $\dfrac{c_1}{c_2} = \dfrac{2}{3}$; these two are never equal regardless of $k$, so the condition can never be fully satisfied. Always check all three ratios, not just two.

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