For the given pair of linear equations,
$a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ .
The values of x and y are given by:

\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}

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\frac{(b_{2} c_{2} - b_{1} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{2} a_{2} - c_{1} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}

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\frac{(b_{2} c_{1} - b_{1} c_{2})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{1} - c_{2} a_{2})}{(a_{1} b_{2} - a_{2} b_{1})}

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Solution

The correct option is A $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}, \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$
The following table is useful to remember the solutions using the method of cross multiplication:

The solution is given by:

$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

$\frac{x}{b_{1} c_{2} - b_{2} c_{1}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

Therefore, $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})} a n d y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ .

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Similar questions

For the given pair of linear equations,
$a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ .

The values of x and y are ( ___, ___ ).

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a_{1} x + b_{1} y + c_{1} = 0

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A pair of linear equations is given by

a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

and

\frac{a_{1}}{a_{2}} \neq \frac{b_{1}}{b_{2}}

In this case, the pair of linear equations is consistent.

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a_{1} x + b_{1} y + c_{1} = 0

and

a_{2} x + b_{2} y + c_{2} = 0

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a_{1} x + b_{1} y + c_{1} = 0

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For the given pair of linear equations, a1x+b1y+c1=0 and a2x+b2y+c2=0. The values of x and y are given by:

For the given pair of linear equations,
$a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ .
The values of x and y are given by: