In each of the following systems of equations determine whether the system has a unique solution, no solution or infinitely many solutions. In case there is a unique solution, find it:
$x-3 y=3$
$3 x-9 y=2$
GIVEN:
$x-3 y=3$
$3 x-9 y=2$
To find: To determine whether the system has a unique solution, no solution or infinitely many solutions
We know that the system of equations
$a_{1} x+b_{1} y=c_{1}$
$a_{2} x+b_{2} y=c_{2}$
For unique solution
$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$
For no solution
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
For infinitely many solution
$\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$
Here,
$\frac{1}{3}=\frac{3}{9} \neq \frac{3}{2}$
$\frac{1}{3}=\frac{1}{3} \neq \frac{3}{2}$
Since $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$ which means $\frac{1}{3}=\frac{1}{3} \neq \frac{3}{2}$ hence the system of equation has no solution.
Hence the system of equation has no solution