A) \[-10/3\]
B) \[-5/3\]
C) \[2/3\]
D) for all real value except\[\frac{-10}{3}\]
Correct Answer: D
Solution :
| The given equations can be rewritten as \[2x+ky-1=0\] and\[3x-5y-7=0\]. |
| On comparing with \[{{a}_{1}}x+{{b}_{1}}y+{{c}_{1}}=0\]and |
| \[{{a}_{2}}x+{{b}_{2}}y+{{c}_{2}}=0\], we get |
| \[{{a}_{1}}=2,\,{{b}_{1}}=k,\,{{c}_{1}}=-1\]and |
| \[{{a}_{2}}=3,\,{{b}_{2}}=-5,\,{{c}_{2}}=-7\] |
| For unique solution, |
| \[\frac{{{a}_{1}}}{{{a}_{2}}}\ne \frac{{{b}_{1}}}{{{b}_{2}}}\Rightarrow \frac{2}{3}\ne \frac{k}{-5}\Rightarrow k\ne \frac{-10}{3}\] |
| Thus, given lines have a unique solution for all real values of k, except \[\frac{-10}{3}\]. |
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