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For what value of k will …

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For what value of k will the following system of linear equation has no solution 3x+y=1 (2k-1)x+(k-1)y=2k+1
  • 3 answers

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

Puja Sahoo? 6 years, 10 months ago

Topr is always ryt.. i mean honey ji ka answer shi hai.....

Honey ??? 6 years, 10 months ago

For no solution a1/a2 not=b1/b2 uske baad value put kar do
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