Are the following pair of linear equations consistent? Justify your answer,
(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 $2(2 x+6 y)=22$
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.
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