Question

The value of $ k $ for which the equations $ 6x - 2y = 3 $ and $ kx - y = 2 $ have infinitely many solutions is:

• A. $ -3 $
• B. $ 3 $
• C. No value
• D. $ \frac{3}{2} $

Hint:

For two linear equations to have infinitely many solutions, their coefficients must be proportional. Ensure all ratios of corresponding coefficients are equal to confirm dependency.

Answer 1

The Correct Option is C

For a system of linear equations to have infinitely many solutions, the equations must represent the same line. This occurs when the ratios of their coefficients are equal: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Given the equations: \[ \begin{cases} 6x - 2y = 3 \quad \text{(1)}
kx - y = 2 \quad \text{(2)} \end{cases} \] Compute the ratios: \[ \frac{6}{k} = \frac{-2}{-1} = \frac{3}{2} \] Simplify the second ratio: \[ \frac{-2}{-1} = 2 \] Now, set the first ratio equal to 2: \[ \frac{6}{k} = 2 \implies k = \frac{6}{2} = 3 \] Check the third ratio: \[ \frac{3}{2} \neq 2 \] Since the ratios are not equal, there is no value of \( k \) that satisfies the condition for infinitely many solutions. Final Answer \[ \boxed{\text{No value}} \]