Solve the following pairs of equations by cross multiplication rule.
$2 x + 3 y = 46, 3 x + 5 y = 74$

x = 8, y = 10

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x = 10, y = 8

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x = 11, y = 8

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x = 8, y = 11

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Solution

The correct option is B $x = 8, y = 10$
Given system of equations are:
$2 x + 3 y = 46$
$3 x + 5 y = 74$
Now, rearranging the equations, we get:
$2 x + 3 y - 46 = 0$
$3 x + 5 y - 74 = 0$
By cross multiplication method, we know that, for a system of linear equations in $x$ and $y$ , of the form $a_{1} x + b_{1} y + c_{1} = 0$ and $a_{2} x + b_{2} y + c_{2} = 0$ , we have:

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

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

$∴ x = \frac{3 (- 74) - 5 (- 46)}{(2 \times 5) - (3 \times 3)}$
$∴ y = \frac{- 46 (3) - 2 (- 74)}{(2 \times 5) - (3 \times 3)}$
$\Rightarrow x = \frac{- 222 - (- 230)}{10 - 9}$
$\Rightarrow y = \frac{- 138 - (- 148)}{10 - 9}$
$\Rightarrow x = 8$
$\Rightarrow y = 10$

Hence, option A is correct.

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