sove inside Z/3Z the systeme { ((5x+7y =10)),((2x+5y =8)) :}


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Question Number 62439 by mathsolverby Abdo last updated on 21/Jun/19

$s o v e i n s i d e Z / 3 Z t h e s y s t e m e$ ${\begin{cases} 5 x + 7 y = 10 \\ 2 x + 5 y = 8 \end{cases}$
Commented by arcana last updated on 22/Jun/19
$Z/3Z=Z_3 ={0^� ,1^� ,2^� } 0^� ={...,−9,−6,−3,0,3,6,9,...} 1^� ={...,−10,−7,−4,1,4,7,10,...} 2^� ={...,−11,−8,−5,2,5,8,11,...} 2^� x+1^� y=1^� 2^� x+2^� y=2^� ⇒2^� y−y=2^� −1^� =1^� y=1^� ⇒2^� x+(1^� )=1^� 2^� x=0^� (cancelativ)⇒x=0^� cause (Z_3 ,+^� ,∙^� ) is a field$
$Z / 3 Z = Z_{3} = {\bar{0}, \bar{1}, \bar{2}}$ $\bar{0} = {. . ., - 9, - 6, - 3, 0, 3, 6, 9, . . .}$ $\bar{1} = {. . ., - 10, - 7, - 4, 1, 4, 7, 10, . . .}$ $\bar{2} = {. . ., - 11, - 8, - 5, 2, 5, 8, 11, . . .}$ $\bar{2} x + \bar{1} y = \bar{1}$ $\bar{2} x + \bar{2} y = \bar{2}$ $\Rightarrow \bar{2} y - y = \bar{2} - \bar{1} = \bar{1}$ $y = \bar{1} \Rightarrow \bar{2} x + (\bar{1}) = \bar{1}$ $\bar{2} x = \bar{0} (cancelativ) \Rightarrow x = \bar{0} cause (Z_{3}, \bar{+}, \bar{\cdot}) is a field$
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