Question

Find the value of k for which each of the following systems of linear equations has an infinite number of solutions:

$(k-3)x+3y=k,kx+ky=12.$

Answer 1

The given pair of linear equation is :
$(k-3)x+3y=k$
$kx+ky=12$
We can write these Equations as :
$(k-3)x+3y-k=0$…….(1)
$kx+ky-12=0$ ………….(2)

On comparing with General form of a pair of linear equations in two variables x & y is:
$a_{1}x+b_{1}y+c_1=0,a_{2}x+b_{2}y+c_2=0$

$a_1=k-3,b_1=-3,c_1=-k$
$a_2=k,b_2=k,c_2=-12$

$\frac{a_1}{a_2}=\frac{k-3}{k},\frac{b_1}{b_2}=\frac{3}{k},\frac{c_1}{c_2}=\frac{-k}{-12}=\frac{k}{12}$

Given: A pair of linear equations has an infinite solution, if

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

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

Taking the first two terms
$\frac{a_1}{a_2}=\frac{b_1}{b_2}$
$\frac{k-3}{k}=\frac{3}{k}$
$k-3=3$
$k=3+3$
$k=6$

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

$k^2=36$
$k=\sqrt{36}$
$k=6$

Hence, the value of k is 6