**Question:**

Do the following pair of linear equations have no solution? Justify your answer.

(i) 2x + 4y = 3 and 12y + 6x = 6

(ii) x = 2y and y = 2x

(iii) 3x + y – 3 = 0 and 2x + – y = 2

**Solution:**

Condition for no solution $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

(i) Yes, given pair of equations,

$2 x+4 y=3$ and $12 y+6 x=6$

Here, $a_{1}=2, b_{1}=4, c_{1}=-3$

$a_{2}=6, b_{2}=12, c_{2}=-6$

$\therefore$ $\frac{a_{1}}{a_{2}}=\frac{2}{6}=\frac{1}{3}, \frac{b_{1}}{b_{2}}=\frac{4}{12}=\frac{1}{3}$

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

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

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

(ii) No, given pair of equations,

$x=2 y$ and $y=2 x$

or $x-2 y=0$ and $2 x-y=0$

Here, $a_{1}=1, b_{1}=-2, c_{1}=0$

$a_{2}=2, b_{2}=-1, c_{2}=0$

$\therefore$ $\frac{a_{1}}{a_{2}}=\frac{1}{2}$ and $\frac{b_{1}}{b_{2}}=\frac{2}{1}$

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

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

(iii) No, given pair of equations,

$3 x+y-3=0$ and $2 x+\frac{2}{3} y-2=0$

Here, $a_{1}=3, b_{1}=1, c_{1}=-3$,

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

$\therefore$ $\frac{a_{1}}{a_{2}}=\frac{3}{2}, \frac{b_{1}}{b_{2}}=\frac{1}{2 / 3}=\frac{3}{2}$

$\frac{c_{1}}{c_{2}}=\frac{-3}{-2}=\frac{3}{2}$

$\because$ $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=\frac{3}{2}$

Hence, the given pair of linear equations is coincident and having infinitely many solutions.