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.
