x [(2),(1) ]+y [(3),(5) ]+ [((−8)),((−11)) ]=0 find x and y


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Question Number 36003 by rishijagtar@gmail.com last updated on 27/May/18

$x [\begin{matrix} 2 \\ 1 \end{matrix}] + y [\begin{matrix} 3 \\ 5 \end{matrix}] + [\begin{matrix} - 8 \\ - 11 \end{matrix}] = 0$ $f i n d x a n d y$
	Answered by tanmay.chaudhury50@gmail.com last updated on 27/May/18

	$2 x + 3 y - 8 = 0$ $x + 5 y - 11 = 0$ $2 x + 3 y = 8$ $2 x + 10 y = 22$ $s o 7 y = 14$ $y = 2$ $x + 10 - 11 = 0$ $x = 1$
	Answered by Rio Mike last updated on 27/May/18

	$(\begin{matrix} 2 x \\ x \end{matrix}) + (\begin{matrix} 3 y \\ 5 y \end{matrix}) + (\begin{matrix} - 8 \\ - 11 \end{matrix}) = 0$ $2 x + 3 y - 8 = 0$ $2 x + 3 y = 8 . . . . . . . . . . . (1)$ $\Rightarrow x + 5 y - 11 = 0$ $x + 5 y = 11 . . . . . . . . . . (2)$ $bring (1) and (2) together$ $1 ∣ 2 x + 3 y = 8 . . . . . . (1)$ $2 ∣ x + 5 y = 11 . . . . . . . (2)$ $2 x + 3 y = 8 . . . . . . . (3)$ $2 x + 10 y = 22 . . . . . . (4)$ $- 7 y = - 14$ $y = 2$ $substitude y = 2 in (1)$ $2 x + 6 = 8$ $2 x = 2$ $x = 1$ $hence p (x, y) \to (1, 2)$
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