Show that the system $2 x + 3 y - 1 = 0, 4 x + 6 y - 4 = 0$ has no solution.

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Solution

$2 x + 3 y - 1 = 0$
$4 x + 6 y - 4 = 0$

For no solution we must have $\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} \neq \frac{c_{1}}{c_{2}}$

Given equation are in the form of
$a_{1} x + b_{1} y + c_{1} = 0, a_{2} x + b_{2} y + c_{2} = 0$

then
$a_{1} = 2, b_{1} = 3, c_{1} = - 1$
$a_{2} = 4, b_{2} = 6, c_{2} = - 4$

$\frac{a_{1}}{a_{2}} = \frac{2}{4} = \frac{1}{2}$
$\frac{b_{1}}{b_{2}} = \frac{3}{6} = \frac{1}{2}$
$\frac{c_{1}}{c_{2}} = \frac{- 1}{- 4} = \frac{1}{4}$

hence
$\frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{1}{2} \neq \frac{c_{1}}{c_{2}}$
hence the given system has no solution

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Similar questions

Show graphically that each of the following given systems of equations is inconsistent.i.e., has no solution:

$2 x + 3 y = 4, 4 x + 6 y = 12.$

2 x + 3 y = 6 4 x + 6 y = 12

3 x - 4 y - 1 = 0

2 x - \frac{8}{3} y + 5 = 0

Show graphically that each of the following given systems of equations has infinitely many solutions:

$2 x + 3 y = 6, 4 x + 6 y = 12.$

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