# Do the following pair of linear equations have no solution? Justify your answer.$2 x+4 y=3$$12 y+6 x=6$

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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.

Updated on 10-Oct-2022 13:27:13