# Are the following pair of linear

Question:

(i) $-3 x-4 y=12$ and $4 y+3 x=12$

(ii) $\frac{3}{5} x-y=\frac{1}{2}$ and $\frac{1}{5} x-3 y=\frac{1}{6}$

(iii) $2 a x+b y=a$ and $4 a x+2 b y-2 a=0 ; a, b \neq 0$

(iv) $x+3 y=11$ and $2(2 x+6 y)=22$

Solution:

Conditions for pair of linear equations are consistent

$\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}$ [unique solution]

and $\frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}$ [infinitely many solutions]

(I) No, the given pair of linear equations

$-3 x-4 y \neq 12$ and $3 x+4 y=12$

Here, $a_{1}=-3, b_{1}=-4, c_{1}=-12 ;$

$a_{2}=3, b_{2}=4, c_{2}=-12$

Now, . $\frac{a_{1}}{a_{2}}=-\frac{3}{3}=-1, \frac{b_{1}}{b_{2}}=-\frac{4}{4}=-1, \frac{c_{1}}{c_{2}}=\frac{-12}{-12}=1$

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

Hence, the pair of linear equations has no solution, i.e., inconsistent.

(ii) Yes, the given pair of linear equations

$\frac{3}{5} x-y=\frac{1}{2}$ and $\frac{1}{5} x-3 y=\frac{1}{6}$

Here, $a_{1}=\frac{3}{5}, b_{1}=-1_{1} c_{1}=-\frac{1}{2}$

and $a_{2}=\frac{1}{5}, b_{2}=-3, c_{2}=-\frac{1}{6}$

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

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

(iii) Yes, the given pair of linear equations

$2 a x+b y-a=0$

and $\quad 4 a x+2 b y-2 a=0, a, b \neq 0$

Here, $\quad a_{1}=2 a, b_{1}=b, c_{1}=-a_{i}$

$a_{2}=4 a, b_{2}=2 b, c_{2}=-2 a$

Now, $\frac{a_{1}}{a_{2}}=\frac{2 a}{4 a}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{b}{2 b}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{-a}{-2 a}=\frac{1}{2}$

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

Hence, the given pair of linear equations has infinitely many solutions; i.e., consistent or dependent.

(iv) No, the given pair of linear equations

$x+3 y=11$ and $2 x+6 y=11$

Here, $a_{1}=1, b_{1}=3, c_{1}=-11$ ...(i)

$a_{2}=2, b_{2}=6, c_{2}=-11$

Now, $\frac{a_{1}}{a_{2}}=\frac{1}{2}, \frac{b_{1}}{b_{2}}=\frac{3}{6}=\frac{1}{2}, \frac{c_{1}}{c_{2}}=\frac{-11}{-11}=1$

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

Hence, the pair of linear equation have no solution $i . e .$, inconsistent.