Question

for what value of k the system of equations K X + 3 Y=1, 12 x + k y= 2 has no solution

Answer 1

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Answer 2

Answer:

$k\neq 6$

Step-by-step explanation:

Given: System of equation $kx+3y=1$ and $12x+ky=2$.

To find: value of k for which system has no solution.

Condition for no solution for system of equation:

$\begin{aligned}&a_{1} x+b_{1} y+c_{1}=0 \\&a_{2} x+b_{2} y+c_{2}=0\end{aligned}$

Condition is: $\left(a_{1} / a_{2}\right)=\left(b_{1} / b_{2}\right) \neq\left(c_{1} / c_{2}\right)$

Here, $a_{1}=k, b_{1}=3,c_{1}= -1,a_{2}=1,b_{2}=k,c_{2}=-2$

So,

$\frac{k}{1}=\frac{3}{k}\neq \frac{-1}{-2} \\k=\frac{1}{2} and k\neq 6$

Therefore, for $k\neq 6$ system has no solution.

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