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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 $-6$.