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

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Detailed Solution

The system of equations

3 x - 4 y = 12

and

6 x - 8 y = 48

has no solution.
As general form of equation,

ax + by + c = 0

Given,

3 x - 4 y = 12

\Rightarrow a_{1} = 3, b_{1} = - 4, c_{1} = 12

…(1)
Therefore,

6 x - 8 y = 48

\Rightarrow a_{2} = 6, b_{2} = - 8, c_{2} = 48

…(2)
Therefore,
If the system of linear equation is in the form of

\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{3}} \neq \frac{c_{1}}{c_{2}}

,
then the pair of linear equations has no solution.
Therefore, from (1) and (2),

\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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