Prove that there is a value of c(≠ 0) for which the system has infinitely many solutions. Find this value.

GIVEN:

$6 x+3 y=c-3$

$12 x+c y=c$

To find: To determine for what value of c the system of equation has infinitely many solution 

We know that the system of equations

$a_{1} x+b_{1} y=c_{1}$

$a_{2} x+b_{2} y=c_{2}$

For infinitely many solution

$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$

Here

$\frac{6}{12}=\frac{3}{c}=\frac{c-3}{c}$

Consider the following

$\frac{6}{12}=\frac{3}{c}$

$c=\frac{12 \times 3}{6}$

$c=6$

Now consider the following for c

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

$3 c=c(c-3)$

$3 c=c^{2}-3 c$

$6 c=c^{2}$

$6 c=c^{2}$

$c=0,6$

But it is given that $c \neq 0 .$ Hence $c=6$

Hence for $c=6$ the system of equation have infinitely many solutions.