Question

For what value of $K$ will the following pair of linear equations have infinitely many solutions?

$Kx+3y-(K-3)=0$
$12x+Ky-K=0$

Answer 1

Given

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

Comparing with $a_{1}x+b_{1}y+c1=0$

$\therefore a_1=k,b_1=3,c=-(k-3)$

$12x+ky-k=0$

Comparing with $a_{1}x+b_{1}y+c1=0$

$\therefore a_1=12,b_1=k,c=-k$

Since equation has infinite number of solutions

So, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$

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

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

$k^2=12\times 3$

$k^2=36$

$k=\pm 6$

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

$3k=k(k-3)$

$3k=k^2-3k$

$k^2-3k-3k=0$

$k^2-6k=0$

$k(k-6)=0$

$k=0,6$

Therefore, $k=6$ satisfies both equations

Hence, $k=6$