MCQ
The system $x - 2y = 3 $ and $3x + ky = 1$ has a unique solution only when :
- A$\text{k}=-6,$
- ✓$\text{k}\neq-6$
- C$\text{k}=0$
- D$\text{k}\neq0$
✓
Answer
Correct option: B.
$\text{k}\neq-6$
$x - 2y = 3$ and $3x + ky = 1$
We know that, the system of linear equations $a_1 x+b_1 x+c_1=0, $
$a_2 x+b_2 y+c_2=0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{1}{3}\neq\frac{-2}{\text{k}}$
$\Rightarrow\text{k}\neq-6.$
We know that, the system of linear equations $a_1 x+b_1 x+c_1=0, $
$a_2 x+b_2 y+c_2=0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}\neq\frac{\text{b}_1}{\text{b}_2}.$
So, $\frac{1}{3}\neq\frac{-2}{\text{k}}$
$\Rightarrow\text{k}\neq-6.$
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