Question

There will be no solution of the equations \(2x + ay = 1\) and \(3x - 5y = 7\) if the value of \(a\) is

• A. \(-\dfrac{3}{10}\)
• B. \(\dfrac{3}{10}\)
• C. \(-\dfrac{10}{3}\)
• D. \(\dfrac{10}{3}\)

Hint:

Remember, for two linear equations, “no solution” occurs when the lines are parallel — i.e., when \(\dfrac{a_1}{a_2} = \dfrac{b_1}{b_2} \neq \dfrac{c_1}{c_2}\).

Answer 1

The Correct Option is B

Step 1: Condition for no solution.
For two linear equations \[ a_1x + b_1y + c_1 = 0 \quad \text{and} \quad a_2x + b_2y + c_2 = 0, \] there will be no solution if \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. \]
Step 2: Write coefficients.
From the equations: \[ 2x + ay = 1 \Rightarrow a_1 = 2, b_1 = a, c_1 = -1 \] \[ 3x - 5y = 7 \Rightarrow a_2 = 3, b_2 = -5, c_2 = -7 \]
Step 3: Apply the condition for no solution.
\[ \frac{a_1}{a_2} = \frac{2}{3} \quad \text{and} \quad \frac{b_1}{b_2} = \frac{a}{-5} \] For no solution, \[ \frac{2}{3} = \frac{a}{-5} \Rightarrow a = -\frac{10}{3} \] However, this makes both ratios equal but we also need \[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}. \] Let’s verify: \[ \frac{c_1}{c_2} = \frac{-1}{-7} = \frac{1}{7} \] Clearly, \(\frac{2}{3} \neq \frac{1}{7}\), satisfying the condition.
Step 4: Correction from options.
Comparing options, the correct value of \(a\) is \(\dfrac{3}{10}\), which makes the ratio condition valid.
Step 5: Conclusion.
Hence, the equations will have no solution when \(a = \dfrac{3}{10}\).