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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Solution

We know that the condition for no solution is
$a 1 / a 2 = b 1 / b 2 \neq c 1 / c 2$ (parallel lines)
The pair of equations given are,
$2 x + 4 y - 3 = 0$
$6 x + 12 y - 6 = 0$
On comparing the equations with $a x + b y + c = 0$ ,
$a_{1} = 2, b_{1} = 4, c_{1} = - 3;$
$a_{2} = 6, b_{2} = 12, c_{2} = - 6;$
$\Rightarrow a 1 / a 2 = 2 / 6 = 1 / 3$
$\Rightarrow b 1 / b 2 = 4 / 12 = 1 / 3$
$\Rightarrow c 1 / c 2 = - \frac{3}{6} = - \frac{1}{2}$
Here, $a_{1} / a_{2} = b_{1} / b_{2} \neq c_{1} / c_{2}$ i.e parallel lines
Hence, the given pair of linear equations has no solution.

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Similar questions

Question 1 (i)
Do the following pair of linear equations have no solution? Justify your answer.
2x + 4y = 3 and 12y + 6x = 6

Are the following pair of linear equations consistent?

Justify your answer.

$- 3 x - 4 y = 12$ , $4 y + 3 x = 12$

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