For what value of \(k\), the system of equations \(3x-ky-3=0\) and \(2x-3y-4=0\) has no solution?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 9/2
Two linear equations \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) have no solution when their lines are parallel and distinct, that is, when \(\displaystyle\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\).
\(\displaystyle\frac{a_1}{a_2}=\frac{3}{2}\) and \(\displaystyle\frac{b_1}{b_2}=\frac{-k}{-3}=\frac{k}{3}\)
Parallel lines: \(\displaystyle\frac{k}{3}=\frac{3}{2}\), so \(\displaystyle k=\frac{9}{2}\)
\(\displaystyle\frac{c_1}{c_2}=\frac{-3}{-4}=\frac{3}{4}\), which is not \(\frac{3}{2}\), so the lines are distinct and never meet