Question

Find the value of k for which the given system of equations has no solution.$kx+3y=k-3;12x+ky=k$

• A. $k=-5$
• B. $k=-2$
• C. $k=-3$
• D. None of these

Answer 1

Answer: (D)

None of these

A system of equations, $a_{1}x+b_{1}y+c_1=0$ and $a_{2}x+b_{2}y+c_2=0$ has no solution for the following condition:
$\frac{a_1}{a_2}=\frac{b_1}{b_2}\ne \frac{c_1}{c_2}$

The equations in the given question are:
$kx+3y-k+3=0$ and $a_1=k,b_1=3,c_1=3$
$12x+ky-k=0$ and $a_2=12,b_2=k,c_2=-k$

Substituting the above values in the condition mentioned above for no solution:

$\frac{a_1}{b_1}=\frac{a_2}{b_2}$

$\frac{k}{12}=\frac{3}{k}$

$k^2=36$
$\Rightarrow k=6$