​Select the option that is true regarding the following labelled Assertion (A) and Reason (R). ​

Assertion (A)	Reason (R)
The system of equations: $$9x+6y=11$$​ and $$7x+ky=9$$​ has no solution, if k = $$\frac{14}{3}$$​.	System of equations $$ax+by=c$$​ $$dx+ey=f$$​ has no solution, if $$\frac{a}{d}=\frac{b}{e}\#\frac{c}{f}$$​

Correct option is A

Given:

Assertion (A):

The system of equations
9x + 6y = 11 and
7x + ky = 9
has no solution if k = 14/3.

Reason (R):

A system of linear equations:
ax + by = c
dx + ey = f
has no solution if  $$\frac{a}{d} = \frac{b}{e} \ne \frac{c}{f} \\[5pt]$$

Solution:

​Analysis:
Compare both equations to general form:

Equation 1: 9x + 6y = 11 → a = 9, b = 6, c = 11

Equation 2: 7x + ky = 9 → d = 7, e = k, f = 9

Let’s apply the condition:

​$$\frac{a}{d} = \frac{b}{e} \Rightarrow \frac{9}{7} = \frac{6}{k} \Rightarrow 9k = 42 \Rightarrow k = \frac{42}{9} = \frac{14}{3} \\[10pt]\text{So if } \mathbf{k = \frac{14}{3}}, \text{ the equations become:} \\\bullet\quad \frac{a}{d} = \frac{b}{e} \neq \frac{c}{f} \text{ condition \textbf{does not hold}, because:} \\[5pt]\frac{9}{7} = \frac{6}{\frac{14}{3}} = \frac{18}{14} = \frac{9}{7}, \quad \text{but} \quad \frac{11}{9} \ne \frac{9}{7}$$​​

So, the lines are parallel but not coincident, hence no solution — this confirms the assertion.

Conclusion:

• Assertion is True (No solution when k=14/3)
• Reason is True (Correct condition for inconsistency)
• Reason correctly explains Assertion

Final Answer: (1) Both A and R are true and R is a correct explanation of A.

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