A) 6
B) 4
C) 5
D) 2
Correct Answer: C
Solution :
[c] The given system of equations is |
\[kx-y-4=0\] |
\[10x-2y-3=0\] |
Here, \[{{a}_{1}}=k,\] \[{{b}_{1}}=-1,\] \[{{c}_{1}}=-4\] and \[{{a}_{1}}=10,\,{{b}_{2}}=-2,\,{{c}_{2}}=-3\] |
The given system will have no solution, if |
\[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\ne \frac{{{c}_{1}}}{{{c}_{2}}}\,\,\,\Rightarrow \,\,\frac{k}{10}=\frac{-1}{-2}\ne \frac{-2}{-3}\] |
\[\Rightarrow \,\,\,\,\,\,\frac{k}{10}=\frac{-1}{-2}\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \,\,\,\,\,\,k=\frac{10}{2}\] |
\[\Rightarrow \,\,\,\,\,\,k=5\] |
Hence, the given system will have no solution, if \[k=5.\] |
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