For what value of k will …

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Ram Kushwah 6 years, 10 months ago
3x+y-1 =0
(2k-1)x+(k-1)y-2k-1 =0
{tex}\begin{array}{l}\frac{\mathrm a1}{\mathrm a2}=\frac{\mathrm b1}{\mathrm b2}\neq\frac{\mathrm c1}{\mathrm c2}\\\frac3{2\mathrm k-1}=\frac1{\mathrm k-1}\\3\mathrm k-3=2\mathrm k-1\\\mathrm k=2\\\mathrm{Now}\;\frac{\mathrm c1}{\mathrm c2}=\frac{-1}{-2\mathrm k-1}=\frac15\neq\frac{\mathrm a1}{\mathrm a2}(=\frac33=1)\end{array}{/tex}
Hence for k=1 the eqns has no solution
1Thank You