MCQ
The system $x + 2y = 3$ and $5x + ky + 7 = 0$ has no solution, when :
- ✓$\text{k}=10$
- B$\text{k}\neq10$
- C$\text{k}=\frac{-7}{3}$
- D$\text{k}=-21$
✓
Answer
Correct option: A.
$\text{k}=10$
$x + 2y = 3$ and $5x + ky + 7 = 0$
$x + 2y - 3 = 0$ and $5x + ky + 7 = 0$
We know that,
the system of linear equations $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, $\frac{1}{5}=\frac{2}{\text{k}}\neq\frac{-3}{7}$
$\Rightarrow\frac{1}{5}=\frac{2}{\text{k}}$
$\Rightarrow\text{k}=10.$
$x + 2y - 3 = 0$ and $5x + ky + 7 = 0$
We know that,
the system of linear equations $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ has a unique solution if $\frac{\text{a}_1}{\text{a}_2}=\frac{\text{b}_1}{\text{b}_2}\neq\frac{\text{c}_1}{\text{c}_2}.$
So, $\frac{1}{5}=\frac{2}{\text{k}}\neq\frac{-3}{7}$
$\Rightarrow\frac{1}{5}=\frac{2}{\text{k}}$
$\Rightarrow\text{k}=10.$
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