The system of equations $3 x - 4 y = 12$ and $6 x - 8 y = 48$ has

2

solution

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1

solution

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Infinite number of solutions

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no solution

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Solution

The correct option is D no solution
$3 x - 4 y = 12$ $\Rightarrow a_{1} = 3, b_{1} = - 4, c_{1} = 12$
$6 x - 8 y = 48$ $\Rightarrow a_{2} = 6, b_{2} = - 8, c_{2} = 48$
$\frac{a_{1}}{a_{2}} = \frac{3}{6} = \frac{1}{2}$
$\frac{b_{1}}{b_{2}} = \frac{- 4}{- 8} = \frac{1}{2}$
$\frac{c_{1}}{c_{2}} = \frac{12}{48} = \frac{1}{4}$
$∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$
The given system of linear equations are inconsistent
Therefore, the given system of equations has no solution.

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