Correct option is (C) x – 2y = 3, 3x – 2y = 1

Condition for unique solution is

\(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)           ____________(1)

We have to check condition (1) for all the given options.

(A) \(\frac{a_1}{a_2}=\frac36=\frac12,\)

\(\frac{b_1}{b_2}=\frac12\)

and \(\frac{c_1}{c_2}=\frac{-2}{-3}=\frac23\)

\(\because\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)

Therefore, the system has no solution.

(B) \(\frac{a_1}{a_2}=\frac26=\frac13,\)

\(\frac{b_1}{b_2}=\frac{-5}{-15}=\frac13\)

and \(\frac{c_1}{c_2}=\frac{-3}{-9}=\frac13\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)

Therefore, this system has infinite many solutions.

(C) \(\frac{a_1}{a_2}=\frac13,\)

\(\frac{b_1}{b_2}=\frac{-2}{-2}=1\)

\(\because\) \(\frac13\neq1\)

\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)

Therefore, this system has unique solution.

(D) \(\frac{a_1}{a_2}=\frac26=\frac13,\)

\(\frac{b_1}{b_2}=\frac{5}{15}=\frac13\)

\(\frac{c_1}{c_2}=\frac{-7}{-3}=\frac73\)

\(\because\) \(\frac13\neq\frac73\)

\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)

Therefore, this system has no solution.

Correct option is C) x – 2y = 3, 3x – 2y = 1