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# For which value(s) of $ k $ will the pair of equations

$ k x+3 y=k-3 $

$ 12 x+k y=k $

have no solution?

Given:

The given system of equations is:

\( k x+3 y=k-3 \)

\( 12 x+k y=k \)

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:

\( k x+3 y-(k-3)=0 \)

\( 12 x+k y-k=0 \)

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

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 $a_2=12, b_2=k, c_2=-k$

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}} \ $

Therefore,

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

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

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

$k^2=36$

$k=\sqrt{36}$

$k=\pm 6$

$\frac{3}{k} ≠ \frac{k-3}{k}$

$k-3≠3$

$k≠3+3$

$k≠6$

This implies,

$k=-6$

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

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