Find the value of $k$ for which the following system of equations has no solution:
$kx\ +\ 3y\ =\ k-\ 3$$12x\ +\ ky\ =\ 6$

Given: 

The given system of equations is:

$kx\ +\ 3y\ =\ k-\ 3$

$12x\ +\ ky\ =\ 6$

To do: 

We have to find the value of $k$ for which the given system of equations has no solution.

Solution:

The given system of equations can be written as:

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

$12x\ +\ ky\ -\ 6=0$

The standard form of system of equations of two variables is $a_{1} x+b_{1} y+c_{1}=0$ and $a_{2} x+b_{2} y-c_{2}=0$.

The condition for which the above system of equations has no solution is

$\frac{a_{1}}{a_{2}} \ =\frac{b_{1}}{b_{2}} ≠ \frac{c_{1}}{c_{2}} \ $

Comparing the given system of equations with the standard form of equations, we have,

$a_1=k, b_1=3, c_1=-(k-3)$ and $a_2=12, b_2=k, c_2=-6$

Therefore,

$\frac{k}{12}=\frac{3}{k}≠\frac{-(k-3)}{-6}$

$\frac{k}{12}=\frac{3}{k}≠\frac{k-3}{6}$

$\frac{k}{12}=\frac{3}{k}$ and $\frac{k}{12}≠\frac{k-3}{6}$

$k\times k=12\times3$ and $k\times6≠12\times(k-3)$

$k^2=36$ and $6k≠12k-36$

$k=\sqrt{36}$ and $6k-12k≠-36$

$k=6$ or $k=-6$ and $-6k≠-36$

$k=6$ or $k=-6$ and $k≠6$

Therefore, $k=6$

The value of $k$ for which the given system of equations has no solution is $6$.  

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Updated on: 10-Oct-2022

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