When using cross multiplication method for solving

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

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

$x i s g i v e n b y \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

True

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False

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Solution

The correct option is A
True

Let

$a_{1} x$ + $b_{1} y$ + $c_{1}$ = 0 ......(1)

$a_{2} x$ + $b_{2} y$ + $c_{2}$ = 0 ......(2)

To solve for 'x' in cross multiplication method
we multiply Eq(1) with $b_{2}$ and Eq(2) with $b_{1}$

we get,

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

Hence the given statement is true.

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Similar questions

When using cross multiplication for solving

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

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

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

When using cross multiplication method for solving

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

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

y is given by $\frac{c_{1} a_{2} - c_{2} a_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

When using cross multiplication for solving

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

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

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

Solve the two equations using method of elimination.

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

In cross multiplication method of solving
$a_{1} x + b_{1} y + c_{1} = 0 a n d a_{2} x + b_{2} y + c_{2} = 0, x = \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$

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When using cross multiplication method for solving a1x + b1y + c1 = 0 a2x + b2y + c2 = 0 x is given by b1c2−b2c1a1b2−a2b1

The correct option is A True Let a1x + b1y + c1 = 0 ......(1) a2x + b2y + c2 = 0 ......(2) To solve for 'x' in cross multiplication method we multiply Eq(1) with b2 and Eq(2) with b1 we get, ⇒x=b1c2−b2c1a1b2−a2b1. Hence the given statement is true.

When using cross multiplication method for solving

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

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

$x i s g i v e n b y \frac{b_{1} c_{2} - b_{2} c_{1}}{a_{1} b_{2} - a_{2} b_{1}}$