Do the following equations represent

Condition for coincident lines,

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

(i) No, given pair of linear equations

$3 x+\frac{y}{7}-3=0$

and $\quad 7 x+3 y-7=0$,

where, $\quad a_{1}=3, b_{1}=\frac{1}{7}, c_{1}=-3$;

$a_{2}=7, b_{2}=3, c_{2}=-7$

Now, $\frac{a_{1}}{a_{2}}=\frac{3}{7}, \frac{b_{1}}{b_{2}}=\frac{1}{21}, \frac{c_{1}}{c_{2}}=\frac{3}{7}$ $\left[\because \frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}\right]$

Hence, the given pair of linear equations has unique solution.

(ii) Yes, given pair of linear equations

$-2 x-3 y-1=0$ and $6 y+4 x+2=0$

where, $a_{1}=-2, b_{1}=-3, c_{1}=-1$

$a_{2}=4, b_{2}=6, c_{2}=2$

Now,   $\frac{a_{1}}{a_{2}}=-\frac{2}{4}=-\frac{1}{2}$

$\frac{b_{1}}{b_{2}}=-\frac{3}{6}=-\frac{1}{2}, \frac{c_{1}}{c_{2}}=-\frac{1}{2}$

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

Hence, the given pair of linear equations is coincident.

(iii) No, the given pair of linear equations are

$\frac{x}{2}+y+\frac{2}{5}=0$ and $4 x+8 y+\frac{5}{16}=0$

Here, $a_{1}=\frac{1}{2}, b_{1}=1, c_{1}=\frac{2}{5}$

$a_{2}=4, b_{2}=8, c_{2}=\frac{5}{16}$

Now, $\frac{a_{1}}{a_{2}}=\frac{1}{8}, \frac{b_{1}}{b_{2}}=\frac{1}{8}, \frac{c_{1}}{c_{2}}=\frac{32}{25}$

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

Hence, the given pair of linear equations has no solution.