Solve the two equations using method of elimination- a 1 x - b 1 y - c 1-0a 2 x - b 2 y - c 2-0A- x -b1 c2-b2-a1 b2-a2- y -c1 a2-a1 b2-i- -b2 c1-2-a2 b1 B- x - b 2 c 1- b 1 c 2- a 1 b 2- a 2 b 1- y -y -C- x-

The correct option is A x = nbsp; (b_1c_2 b_2c_1)/(a_1b_2 a_2b_1), y = nbsp; (c_1a_2 c_2a_1)/(a_1b_2 a_2b_1) Given two equations are a_1x + b_1y + c_1 = 0 ...

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Byju's Answer

Standard IX

Mathematics

Method of Elimination

Solve the two...

Question

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

The solution (x,y) for the given pair of linear equations is $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0

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 solution $(x, y)$ is

For the pair of linear equations $a_{1} x$ + $b_{1} y$ + $c_{1} = 0$ and $a_{2} x$ + $b_{2} y$ + $c_{2} = 0$ , the solution set are $x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and $y = \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ . They can also be written as:

The solution set x= $\frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ and y= $\frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$ to the pair of linear equations $a_{1}$ x + $b_{1}$ y + $c_{1}$ =0 and $a_{2}$ x + $b_{2}$ y + $c_{2}$ =0 can be written as

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

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Solve the two equations using method of elimination. a1x+b1y+c1=0 a2x+b2y+c2=0

Solve the two equations using method of elimination.

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