For the following system of equation determine the value of $k$ for which the given system of equations has unique solution. $kx+3y=( k-3)$ and $12x+ky=k$.

Given:  The system of equations: $kx+3y=( k-3)$ and $12x+ky=k$.

To do: To find the value of $k$ for which the given system of equations has a unique solution.

Solution:

The given system of equations: $kx+3y=(k-3)$

$\Rightarrow kx+3y-(k-3)=0\ .....( i)$

$12x+ky=k$

$\Rightarrow 12x+ky-k=0\ ..... ( ii)$  

Here, $a_1=k,\ b_1= 3,\ c_1=-(k-3)$ and $a_2=12,\ b_2=k,\ c_2=-k$  

For a unique solution, we must have: 

$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$

$\Rightarrow \frac{k}{12}≠\frac{3}{k}$

$\Rightarrow k^2≠36$

$\Rightarrow k≠\pm\sqrt{36}$

$\Rightarrow k≠\pm 6$

Thus, for all real values of $k$, other than $\pm 6$, the given system of equations will have a unique solution.
 

Updated on: 10-Oct-2022

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