Correct option is (A) 3x – y = 1, 6x – 2y = 5
(A) \(\frac{a_1}{a_2}=\frac36=\frac12,\)
\(\frac{b_1}{b_2}=\frac{-1}{-2}=\frac12\)
and \(\frac{c_1}{c_2}=\frac{-1}{-5}=\frac15\)
\(\because\) \(\frac12\neq\frac15\)
\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\)
Therefore, this system has no solution.
Hence, this system of equations is inconsistent.
(B) \(\frac{a_1}{a_2}=\frac4{12}=\frac13,\)
\(\frac{b_1}{b_2}=\frac{6}{18}=\frac13\)
and \(\frac{c_1}{c_2}=\frac{-7}{-21}=\frac13\)
\(\therefore\) \(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
\(\therefore\) This system has infinitely many solutions.
Hence, this system of equations is consistent.
(C) \(\frac{a_1}{a_2}=\frac49,\)
\(\frac{b_1}{b_2}=\frac98\)
\(\because\) \(\frac49\neq\frac98\)
\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)
\(\therefore\) This system has unique solution.
Hence, this system of equations is consistent.
(D) \(\frac{a_1}{a_2}=\frac49,\)
\(\frac{b_1}{b_2}=\frac{12}{9}=\frac43\)
\(\because\) \(\frac49\neq\frac43\)
\(\therefore\) \(\frac{a_1}{a_2}\neq\frac{b_1}{b_2}\)
\(\therefore\) This system has unique solution.
Hence, this system of equations is consistent.
