Question

Find the value of k for which the given system of equations has a unique solution.

$(k-3)x+3y=k;kx+ky=12$

• A. $k\ne 3$
• B. $k\ne 9$
• C. $k\ne 6$
• D. $k\ne 2$

Answer 1

The correct option is C $k\ne 6$

Comparing $(k-3)x+3y=k$ with $a_{1}x+b_{1}y+c_1=0$,

and $kx+ky=12$ with $a_{2}x+b_{2}y+c_2=0$,

we get, $a_1=(k-3),a_2=k,b_1=3,b_2=k$

Now,

Given system of equations has unique solution
$\Rightarrow \frac{a_1}{a_2}\ne \frac{b_1}{b_2}$
$\frac{(k-3)}{k}\ne \frac{3}{k}$

$\Rightarrow k^2-3k\ne 3k$

$\Rightarrow k^2\ne 6k$

$\Rightarrow k^2-6k\ne 0$

$\Rightarrow k(k-6)\ne 0$

$\Rightarrow k\ne 0$ and $k\ne 6$

So $k\ne 6$ is the correct option among given options.