solve in Z^2 2x+3y =7


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Question Number 42296 by maxmathsup by imad last updated on 22/Aug/18

$s o l v e i n Z^{2} 2 x + 3 y = 7$
Commented by math khazana by abdo last updated on 22/Aug/18

$w e h a v e 2 (- 1) + 3 (3) = 7 s o (- 1, 3) i s a s p e c i a l$ $s o l u t i o n (e) \Rightarrow 2 x + 3 y - 2 (- 1) - 3 (3) = 0 \Rightarrow$ $2 (x + 1) + 3 (y + 3) = 0 \Rightarrow 2 (x + 1) = - 3 (y + 3) \Rightarrow$ $3 d i v i d e 2 (x + 1) b u t D (2, 3) = 1 \Rightarrow 3 d i v i d e x + 1 \Rightarrow$ $x + 1 = 3 k \Rightarrow x = 3 k - 1 \Rightarrow 3 y = 7 - 2 x$ $= 7 - 2 (3 k - 1) = 7 - 6 k + 2 = 9 - 6 k \Rightarrow$ $y = 3 - 2 k f i n a l l y x = 3 k - 1 a n d y = - 2 k + 3 w i t h$ $k \in Z .$
	Answered by ajfour last updated on 22/Aug/18

	$2 (x + y) + y = 7$ $l e t x + y = n$ $\Rightarrow 2 n + y = 7$ $\Rightarrow y = 7 - 2 n$ $x = 3 n - 7 (n \in Z) .$
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