The value of k, for which the system of equations \(3x-ky-20=0\), and \(6x-10y+40=0\) has no solution, is:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 5
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 \(\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}\ne\frac{c_1}{c_2}\).
\(\displaystyle \frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}\)
\(\displaystyle \frac{b_1}{b_2}=\frac{-k}{-10}=\frac{k}{10}\)
\(\displaystyle \frac{c_1}{c_2}=\frac{-20}{40}=-\frac{1}{2}\)
Parallel lines need \(\displaystyle \frac{k}{10}=\frac{1}{2}\), so \(k=5\); and \(\frac{1}{2}\ne-\frac{1}{2}\), so the lines are not the same line and never meet