If 3x - ky + 8 = 0 and 9x - 18y + 20 = 0 have no solution, find k.
6
For two lines \(a_1x+b_1y+c_1=0\) and \(a_2x+b_2y+c_2=0\) to have no solution, they must be parallel but distinct, i.e. \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}\ne\dfrac{c_1}{c_2}\).
Here the system is \(3x-ky+8=0\) and \(9x-18y+20=0\), so \(a_1=3,\ b_1=-k,\ c_1=8\) and \(a_2=9,\ b_2=-18,\ c_2=20\).
Set the coefficient ratios equal: \(\dfrac{3}{9}=\dfrac{-k}{-18}\).
Simplify: \(\dfrac{1}{3}=\dfrac{k}{18}\), so \(k=\dfrac{18}{3}=6\).
Check the constant ratio is different: \(\dfrac{c_1}{c_2}=\dfrac{8}{20}=\dfrac{2}{5}\), which is not \(\dfrac{1}{3}\), confirming the lines are parallel and truly have no common solution.
The key condition is the parallel-lines (no-solution) criterion for a pair of linear equations.
Hence the required value is k = 6.
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