For the pair of linear equations a1x - b1y - c1-0 and a2x - b2y - c2-0- the solution set are x - b1c2-b2c1-a1b2-a2b1 and y - c1a2-c2a1-a1b2-a2b1- They can also be written as-

We have x = b_1c_2 b_2c_1/(a_1b_2 a_2b_1 and y = c_1a_2 c_2a_1/a_1b_2 a_2b_1 We can write this as x/b_1c_2 b_2c_1 = y/c_1a_2 c_2a_1 = 1/a_1b_2 ...

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Byju's Answer

Standard X

Mathematics

Classification of Pair of Linear Equations

For the pair ...

Question

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:

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

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

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

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

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Solution

The correct option is B
$\frac{x}{b_{1} c_{2} - b_{2} c_{1}}$ = $\frac{y}{c_{1} a_{2} - c_{2} a_{1}}$ = $\frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

We have $x = \frac{b_{1} c_{2} - b_{2} c_{1}}{(a_{1} b_{2} - a_{2} b_{1}}$ and $y = \frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

We can write this as $\frac{x}{b_{1} c_{2} - b_{2} c_{1}}$ = $\frac{y}{c_{1} a_{2} - c_{2} a_{1}}$ = $\frac{1}{a_{1} b_{2} - a_{2} b_{1}}$

This is the basis for the cross multiplication method.

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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

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 .

Solve the two equations using method of elimination.

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

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For the pair of linear equations a1x + b1y + c1=0 and a2x + b2y + c2=0, the solution set are x=(b1c2−b2c1)(a1b2−a2b1) and y=(c1a2−c2a1)(a1b2−a2b1). They can also be written as:

The correct option is B xb1c2−b2c1 = yc1a2−c2a1 = 1a1b2−a2b1 We have x=b1c2−b2c1(a1b2−a2b1 and y=c1a2−c2a1a1b2−a2b1 We can write this as xb1c2−b2c1 = yc1a2−c2a1 = 1a1b2−a2b1 This is the basis for the cross multiplication method.