The solution x- y for the given pair of linear equations is a1 x-b1 y-c1-0 and a2 x-b2 y-c2-0A- None of theseB- b1 c2-b2 c1 -a1 b2-a2 b1-c1 a2-c2 a1 -a1 b2-a2 b1C- b2 C 2-b1 c 1 - a 1 b 2- a 2 b 1- c 2 a 2- c 1 a 1 - a 1 b 2- a 2 b 1D- b2 c1-b1 c2 -a1 b2-a2 b1-c1 a1-c2 a2 -a1 b2-a2 b1

The correct option is B ((b1c2 b2c1)/(a1b2 a2b1), (c1a2 c2a1)/(a1b2 a2b1)) Let us use elimination method to solve the given pair of equations Given two equatio ...

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Standard X

Mathematics

Elimination Method of Finding Solution of a Pair of Linear Equations

The solution ...

Question

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

((b₁c₂-b₂c₁)/(a₁b₂-a₂b₁), (c₁a₂-c₂a₁)/(a₁b₂-a₂b₁))

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((b₂c₂-b₁c₁)/(a₁b₂-a₂b₁), (c₂a₂-c₁a₁)/(a₁b₂-a₂b₁))

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((b₂c₁-b₁c₂)/(a₁b₂-a₂b₁), (c₁a₁-c₂a₂)/(a₁b₂-a₂b₁))

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None of these

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Solution

The correct option is A
((b₁c₂-b₂c₁)/(a₁b₂-a₂b₁), (c₁a₂-c₂a₁)/(a₁b₂-a₂b₁))

Let us use elimination method to solve the given pair of equations

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

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

Multiply first equation by b₂ and second equation by $b_{1}$ , to 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

Subtracting these two equations

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

Substitute this x in any of the given equation we get

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 solution $(x, y)$ is

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 ( ___, ___ ).

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:

Solve the two equations using method of elimination.

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

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

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The solution (x,y) for the given pair of linear equations is a1x + b1y + c1=0 and a2x + b2y + c2=0

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