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Do the following pair of linear equations have no solution? Justify your answer.
\( 2 x+4 y=3 \)
\( 12 y+6 x=6 \)
To find :
We have to find whether the given pairs of equations have no solution.
Solution:
We know that,
The condition for no solution is
$\frac{a_1}{a_2}=\frac{b_1}{b_2}≠\frac{c_1}{c_2}$
(i) \( 2 x+4 y-3=0 \)
\( 12 y+6 x-6=0 \)
Here,
$a_1=2, b_1=4, c_1=-3$
$a_2=6, b_2=12, c_2=-6$
Therefore,
$\frac{a_1}{a_2}=\frac{2}{6}=\frac{1}{3}$
$\frac{b_1}{b_2}=\frac{4}{12}=\frac{1}{3}$
$\frac{c_1}{c_2}=\frac{-3}{-6}=\frac{1}{2}$
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.
(ii) \( x-2 y=0 \)
\( 2 x-y=0 \)
Here,
$a_1=1, b_1=-2, c_1=0$
$a_2=2, b_2=-1, c_2=0$
Therefore,
$\frac{a_1}{a_2}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{-2}{-1}=2$
$\frac{c_1}{c_2}=\frac{-3}{-6}=\frac{1}{2}$
Here,
$\frac{a_1}{a_2}≠\frac{b_1}{b_2}$
Hence, the given pair of linear equations has a unique solution.
(iii) \( 3 x+y-3=0 \)
\( 3(2 x)+3(\frac{2}{3} y)=3(2) \)
$6x+2y-6=0$
Here,
$a_1=3, b_1=1, c_1=-3$
$a_2=6, b_2=2, c_2=-6$
Therefore,
$\frac{a_1}{a_2}=\frac{3}{6}=\frac{1}{2}$
$\frac{b_1}{b_2}=\frac{1}{2}$
$\frac{c_1}{c_2}=\frac{-3}{-6}=\frac{1}{2}$
Here,
$\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
Hence, the given pair of linear equations represent coincident lines.