Solve the two equations using method of elimination.

$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$

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

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

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

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x = $\frac{(b_{1} c_{2} + b_{2} c_{1})}{(a_{1} b_{2} + a_{2} b_{1})}$ , y = $\frac{(c_{1} a_{2} + c_{2} a_{1})}{(a_{1} b_{2} + a_{2} b_{1})}$

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Solution

The correct option is A
x = $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ , y = $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

Given two equations are
$a_{1} x + b_{1} y + c_{1} = 0$
and $a_{2} x + b_{2} y + c_{2} = 0$

On multiplying the first equation by $b_{2}$ and second equation by $b_{1}$ , we get

$b_{2}$ $a_{1} x$ + $b_{2}$ $b_{1} y$ + $b_{2}$ $c_{1}$ = 0

$b_{1}$ $a_{2} x$ + $b_{1}$ $b_{2} y$ + $b_{1}$ $c_{2}$ = 0

On subtracting these two equations, we get

$(b_{2} a_{1} - b_{1} a_{2}) x + (b_{2} b_{1} - b_{1} b_{2}) y + (b_{2} c_{1} - b_{1} c_{2}) = 0$

i.e., $(b_{2}$ $a_{1}$ – $b_{1}$ $a_{2}) x$ = $b_{1}$ $c_{2}$ – $b_{2}$ $c_{1}$

Therefore $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ ,

On substituting the value of x in equation (1), we get

$a_{1} (\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}) + b_{1} y + c_{1} = 0$

$\Rightarrow b_{1} y = - c_{1} - a_{1} (\frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}})$

$\Rightarrow b_{1} y = \frac{- c_{1} (a_{1} b_{2} - a_{2} b_{1}) - a_{1} (b_{1} c_{2} - b_{2} c_{1})}{a_{1} b_{2} - a_{2} b_{1}}$

$\Rightarrow y = \frac{b_{1} \times (c_{1} a_{2} - c_{2} a_{1})}{b_{1} \times (a_{1} b_{2} - a_{2} b_{1})}$

$\Rightarrow y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

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Similar questions

Solve the two equations using method of elimination.

$a_{1} x + b_{1} y + c_{1} = 0$
$a_{2} x + b_{2} y + c_{2} = 0$

a_{1} x + b_{1} y + c_{1} = 0 a_{2} x + b_{2} y + c_{2} = 0

When using cross multiplication method for solving

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

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

$x i s g i v e n b y \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

When using cross multiplication method for solving

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

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

y is given by $\frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

When using cross multiplication for solving

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

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

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

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