Which of the following pair of linear equations has no solution?(A) 2…

The correct answer is Option (2) → (A), (C) and (D) only

For \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\), there is no solution when \(\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\): the two lines have the same slope but are different lines, so they are parallel and never meet.

(A) \(2x+3y-1=0\) and \(4x+6y-4=0\) Correct.

• Ratios of the x and y coefficients: \(\frac{2}{4}\) and \(\frac{3}{6}\), both \(\frac{1}{2}\)

• Ratio of the constants: \(\frac{-1}{-4}=\frac{1}{4}\), different, so the lines are parallel

(B) \(x-2y=0\) and \(3x+4y-20=0\) Incorrect.

• Ratios of the x and y coefficients: \(\frac{1}{3}\) and \(\frac{-2}{4}\), not equal

• So the lines cross and there is one solution (\(x=4\), \(y=2\))

(C) \(x+2y-3=0\) and \(5x+10y+7=0\) Correct.

• Ratios of the x and y coefficients: \(\frac{1}{5}\) and \(\frac{2}{10}\), both \(\frac{1}{5}\)

• Ratio of the constants: \(\frac{-3}{7}\), different, so the lines are parallel

(D) \(x+2y-4=0\) and \(2x+4y-12=0\) Correct.

• Ratios of the x and y coefficients: \(\frac{1}{2}\) and \(\frac{2}{4}\), both \(\frac{1}{2}\)

• Ratio of the constants: \(\frac{-4}{-12}=\frac{1}{3}\), different, so the lines are parallel