MCQ
If the system $6x - 2y = 3, kx - y = 2$ has a unique solution, then :
- A$\text{k}\neq{4}$
- B$\text{k} = 4$
- ✓$\text{k} \neq 3$
- D$\text{k} = 3$
✓
Answer
Correct option: C.
$\text{k} \neq 3$
If the system has a unique solution, then $\frac{\text{a}_{1}}{\text{a}_{2}}\neq\frac{\text{b}_{1}}{\text{b}_{2}}$
Here $a_1=6, a_2=k, b_1=-2$ and $b_2=-1$
$\therefore\frac{6}{\text{k}}\neq\frac{-2}{-1}$
$\Rightarrow {3}\text{k}\neq{6}$
$\Rightarrow\text{k}\neq{3}$
${2}\text{k}\neq{6}, \text{ k}\neq{3}$
Here $a_1=6, a_2=k, b_1=-2$ and $b_2=-1$
$\therefore\frac{6}{\text{k}}\neq\frac{-2}{-1}$
$\Rightarrow {3}\text{k}\neq{6}$
$\Rightarrow\text{k}\neq{3}$
${2}\text{k}\neq{6}, \text{ k}\neq{3}$
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