Solve using cross multiplication method

2x + 3y = 17 and 3x - 2y = 6

(4, 3)

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(3, 4)

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(3, -4)

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(-3, 4)

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Solution

The correct option is A
(4, 3)

Given equations: 2x + 3y = 17 and 3x - 2y = 6

$a_{1} = 2, b_{1} = 3, c_{1} = - 17$ and $a_{2} = 3, b_{2} = - 2, c_{2} = - 6$

x y 1

3 -17 2 3

-2 -6 3 -2

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

$\Rightarrow$ $\frac{x}{- 18 - 34}$ = $\frac{y}{- 51 + 12}$ = $\frac{1}{- 4 - 9}$

$\Rightarrow$ $\frac{x}{- 52}$ = $\frac{y}{- 39}$ = $\frac{1}{- 13}$

$\Rightarrow$ $\frac{x}{- 52}$ = $\frac{1}{- 13}$ $\Rightarrow$ x = 4

$\Rightarrow$ $\frac{y}{- 39}$ = $\frac{1}{- 13}$ $\Rightarrow$ y = 3

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