The solution set x - b 1 c 2- b 2 c 1- a 1 b 2- a 2 b 1 and y - c 1 a 2- c 2 a 1- a 1 b 2- a 2 b 1 to the pair of linear equations given below can be written as-a 1 x - b 1 y - c 1-0 and a 2 x - b 2 y - c 2-0A- None of these B- x - b 2 C 2- b 1 c 1- y - a 2 C 2- a 1 c 1-1 - b 2 a 2- b 1 a 1C- x - b 2 C 1- b 1 c 1- y - a 2 c 1- a 1 c 1-1 - b 2 a 1- b 1 a 2D- x - b 1 C 2- b 2 C 1- y - a 2 c 1- a 1 c 2-1 - b 2 a 1- b 1 a 2

The correct option is D x/b1c2−b2c1 = y/a2c1−a1c2 = 1/b2a1−b1a2 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 W ...

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Byju's Answer

Standard IX

Mathematics

Solving Equations Using Method of Substitution

The solution ...

Question

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 given below can be written as:

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

x/b₂c₂-b₁c₁ = y/a₂c₂-a1c1 = 1/b₂a₂-b₁a₁

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x/b₁c₂−b₂c₁ = y/a₂c₁−a₁c₂ = 1/b₂a₁−b₁a₂

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x/b₂c₁−b₁c₁ = y/a₂c1−a₁c₁ = 1/b₂a₁−b₁a₂

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None of these

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Solution

The correct option is B
x/b₁c₂−b₂c₁ = y/a₂c₁−a₁c₂ = 1/b₂a₁−b₁a₂

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

This is the basis for cross multiplication method.

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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

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:

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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The solution set x= (b1c2−b2c1)(a1b2−a2b1) and y= (c1a2−c2a1)(a1b2−a2b1) to the pair of linear equations given below can be written as: a1x + b1y + c1=0 and a2x + b2y + c2=0 .

The correct option is B x/b1c2−b2c1 = y/a2c1−a1c2 = 1/b2a1−b1a2 We have x = b1c2−b2c1(a1b2−a2b1 and y = c1a2−c2a1a1b2−a2b1 We can write this as xb1c2−b2c1 = ya2c1−a1c2 = 1b2a1−b1a2 This is the basis for cross multiplication method.

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 given below can be written as:

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