Question

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

Answer 1

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

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 $a_1=k-3,b_1=3,c_1=-k$

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

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

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

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

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

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

$\Rightarrow (k=0or,k=6)$ and $(k=\pm 6)$

$\Rightarrow k=6$

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