For the given pair of linear equations- a 1 x - b 1 y - c 1-0 and a 2 x - b 2 y - c 2-0The values of x and y are -A- b2 c2-b1 c1-72- c2 a2-c1 a1-l-p2 c1-a2 b1- c1 a2-c2 a1-a1 b2-a2 b1C- b1 c2-b-a1 b2-c-D- None of the above

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Method of Cross Multiplication

For the given...

Question

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 ( _, _ ).

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

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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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$N o n e o f t h e a b o v e$

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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:

$\frac{x}{b_{1} c_{2} - b 2 C_{2}} = \frac{y}{c_{1} a_{2} - c_{2} a_{1}} = \frac{1}{a_{1} b_{2} - a_{2} b_{1}}$
$Therefore x = \frac{(b_{1} c_{2} - b_{2} c_{1})}{(a_{1} b_{2} - a_{2} b_{1})} and y= \frac{(c_{1} a_{2} - c_{2} a_{1})}{(a_{1} b_{2} - a_{2} b_{1})}$

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and

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.

The values of x and y are ( __ , __ ).
Fill in the blanks.

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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 ( ___, ___ ).

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 ( _, _ ).