Find the values of k for which the system will have (i) a unique solution, and (ii) no solution. Is there a value of k for which the system has infinitely many solutions?

GIVEN:

$2 x+k y=1$

$3 x-5 y=7$

To find: To determine for what value of k the system of equation has

(1) Unique solution 

(2) No solution

(3) Infinitely many solution

We know that the system of equations

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

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

(1) For Unique solution

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$

Here,

$\frac{2}{3} \neq \frac{k}{-5}$

$k \neq \frac{-10}{3}$

Hence for $k \neq \frac{-10}{3}$ the system of equation has unique solution

(2) For no solution

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

Here,

$\frac{2}{3}=\frac{k}{-5} \neq \frac{1}{7}$

$\frac{2}{3}=\frac{k}{-5} \quad$ and $\quad \frac{k}{-5} \neq \frac{1}{7}$

$k=\frac{-10}{3} \quad$ and $\quad k \neq \frac{-5}{7}$

Hence for $k=\frac{-10}{3}$ the system of equation has no solution

(3) For infinitely many solution

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

Here,

$\frac{2}{3}=\frac{k}{-5} \neq \frac{1}{7}$

$\Rightarrow \quad k=\frac{-10}{3}$

But since here $\frac{k}{-5} \neq \frac{1}{7}\left(\right.$ as $\left.k=\frac{-10}{3}\right)$

Hence the system does not have infinitely many solutions.