Question

For what value of ‘K’ will the following pair of linear equations have infinitely many solutions
Kx+3y=k-3; 12x+ky=k
[or kx+3y-k+3=0 ; 12x+ky-k=0]

Answer 1

kx+3y=k-3
12x+ky=k
compare with $a_{1}x+b_{1}y=c_1$
$a_{2}x+b_{2}y=c_2$
$a_1=k,a_2=12,b_1=3,b_2=k$
$c_1=k-3,c_2=k$
For infinite solutions, $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\therefore \frac{k}{12}=\frac{3}{k}=\frac{k-3}{k}$
$\frac{k}{12}=\frac{3}{k}\Rightarrow k^2=36$
$\therefore k=6$.