Question

For the following system of equation determine the value of k for which the given system of equation has infinitely many solution.
$kx+3y=k-3$
$12x+ky=k$

Answer 1

The given system of equation is
$kx+3y-(k-3)=0$
$12x+ky-k=0$

This system of equation is of the form
$a_{1}x+b_{1}y+c_1=0$
$a_{2}x+b_{2}y+c_2=0$

where $a_1=k,b_1=3,c_1=-(k-3)$

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

For infinitely many solutions, we must have
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

$\Rightarrow \frac{k}{12}=\frac{3}{k}$ and $\frac{3}{k}=\frac{k-3}{k}$

$\Rightarrow k^2=36$ and $k^2-3k=3k$

$\Rightarrow k^2=36$ and $k^2-6k=0$

$\Rightarrow (k=\pm 6)$ and $(k=0ork=6)$

$\Rightarrow k=6$

Hence, the given system of equations has infinitely many solutions, if $k=6$.