Question

If the pair of linear equations $2x + 3y = 5$ and $4x + ky = 10$ has infinitely many solutions, then the value of $k$ is:

• A. $\dfrac{1}{2}$
• B. $1$
• C. $3$
• D. $6$

Hint:

For infinitely many solutions, all ratios $\dfrac{a_1}{a_2}$, $\dfrac{b_1}{b_2}$, and $\dfrac{c_1}{c_2}$ must be equal.

Answer 1

The Correct Option is C

Step 1: Recall the condition for infinitely many solutions.
For a pair of linear equations: \[ a_1x + b_1y + c_1 = 0 \quad \text{and} \quad a_2x + b_2y + c_2 = 0, \] the condition for infinitely many solutions is: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. \]
Step 2: Identify coefficients from the given equations.
Given: \[ 2x + 3y = 5 \quad \text{and} \quad 4x + ky = 10 \] So, \[ a_1 = 2, \; b_1 = 3, \; c_1 = -5 \] \[ a_2 = 4, \; b_2 = k, \; c_2 = -10 \]
Step 3: Apply the condition.
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \] Using $\frac{a_1}{a_2} = \frac{c_1}{c_2}$: \[ \frac{2}{4} = \frac{-5}{-10} \Rightarrow \frac{1}{2} = \frac{1}{2} \text{ (satisfied)} \] Now, for infinitely many solutions: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \Rightarrow \frac{2}{4} = \frac{3}{k} \]
Step 4: Solve for $k$.
\[ \frac{1}{2} = \frac{3}{k} \Rightarrow k = 6 \] Wait — check calculation: \[ \frac{2}{4} = \frac{3}{k} \Rightarrow \frac{1}{2} = \frac{3}{k} \Rightarrow k = 6 \] Correction: That gives $k = 6$, not 3. Let's verify properly with constants: \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \Rightarrow \frac{2}{4} = \frac{3}{k} = \frac{5}{10} \Rightarrow \frac{1}{2} = \frac{3}{k} = \frac{1}{2} \Rightarrow k = 6 \] So the correct value of $k$ is $6$.