The equation $2 x - 3 y + 5 = 0$ and $6 y - 4 x + 10 = 0$ when solved simultaneously have

only one solution

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no solution

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only two solutions

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infinite number of solutions

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Solution

The correct option is D infinite number of solutions
The equations are
$\Rightarrow 2 x - 3 y + 5 = 0$
$\Rightarrow a_{1} = 2, b_{1} = - 3, c_{1} = 5$
$\Rightarrow - 4 x + 6 y - 10 = 0$
$\Rightarrow a_{2} = - 4, b_{2} = 6, c_{2} = - 10$
$\frac{a_{1}}{a_{2}} = \frac{2}{- 4} = \frac{- 1}{2}$
$\dfrac{b_{1}}{b_{2}}=\dfrac{-3}{6}=\dfrac{-1}{2}$
$ \dfrac{c_{1}}{c_{2}} =\frac{5}{-10} =\dfrac{-1}{2}$

$\Rightarrow \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}}$
The lines are coincidental, has infinite number of solutions and the system of linear equations is consistent.

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