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Are the following pair of linear equations consistent? Justify your answer.
$ -3 x-4 y=12 $
$ 4 y+3 x=12 $
To find :
We have to find whether the given pairs of linear equations are consistent.
Solution:
We know that,
The condition for consistent pair of linear equations is,
$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$ [For unique solution]
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$ [For infinitely many solutions]
(i) \( -3 x-4 y-12=0 \)
\( 4 y+3 x-12=0 \)
Here,
$a_1=-3, b_1=-4, c_1=-12$
$a_2=4, b_2=3, c_2=-12$
Therefore,
$\frac{a_1}{a_2}=\frac{-3}{4}$
$\frac{b_1}{b_2}=\frac{-4}{3}$
$\frac{c_1}{c_2}=\frac{-12}{-12}=1$
Here,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
Hence, the given pair of linear equations has no solution and therefore inconsistent.
(ii) \( \frac{3}{5} x-y=\frac{1}{2} \)
$10(\frac{3}{5}x)-10(y)=10(\frac{1}{2})$
$6x-10y-5=0$
\( \frac{1}{5} x-3 y=\frac{1}{6} \)
$30(\frac{1}{5}x)-30(3y)=30(\frac{1}{6})$
$6x-90y-5=0$
Here,
$a_1=6, b_1=-10, c_1=-5$
$a_2=6, b_2=-90, c_2=-5$
Therefore,
$\frac{a_1}{a_2}=\frac{1}{1}=1$
$\frac{b_1}{b_2}=\frac{-10}{-90}=\frac{1}{9}$
$\frac{c_1}{c_2}=\frac{-5}{-5}=1$
Here,
$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$
Hence, the given pair of linear equations has unique solution and therefore consistent.
(iii) \( 2 a x+b y-a=0 \)
\( 4 a x+2 b y-2 a=0; a, b ≠ 0 \)
Here,
$a_1=2a, b_1=b, c_1=-a$
$a_2=4a, b_2=2b, c_2=-2a$
Therefore,
$\frac{a_1}{a_2}=\frac{2a}{4a}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{b}{2b}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-a}{-2a}=\frac{1}{2}$
Here,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Hence, the given pair of linear equations has infinitely many solutions and therefore consistent.
(iv) \( x+3 y-11=0 \)
\( 4 x+12 y-22=0 \)
Here,
$a_1=1, b_1=3, c_1=-11$
$a_2=4, b_2=12, c_2=-22$
Therefore,
$\frac{a_1}{a_2}=\frac{1}{4}$
$\frac{b_1}{b_2}=\frac{3}{12}=\frac{1}{4}$
$\frac{c_1}{c_2}=\frac{-11}{-22}=\frac{1}{2}$
Here,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
Hence, the given pair of linear equations has no solution and therefore inconsistent.