Question

Find the value of $k$ for which the given system of equations has infinite number of solutions.

$kx+3y=k-3$ and $12x+ky=k$

Answer 1

For the two lines of the form,

$a_{1}x+b_{1}y=c_1$ and $a_{2}x+b_{2}y=c_2$ ......(1)

The condition for the infinite solution is,

$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ ....(2)

The given equation are $kx+3y=k-3$ and $12x+ky=k$

Comparing these with equations in (1)

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

Putting these values in (2)

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

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

$k^2=36$

$k=+6$ or $-6$

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

$k^2=12k-36$

$k^2+36-12k$=0

$(k-6{)}^2=0$

$k=+6$ or $-6$

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

$3k=k^2=3k$

$k^2=6k$

$k=0$ or $6$

The only value common from (i) ,(ii) and (iii) is $k=6$

Therefore $k=6$