{ ((4x=2(mod 9))),((7x=2 (mod 13))) :}


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Question Number 175320 by cortano1 last updated on 27/Aug/22

${\begin{cases} 4 x = 2 (m o d 9) \\ 7 x = 2 (m o d 13) \end{cases}$
Commented by cortano1 last updated on 27/Aug/22

$i g o t x = 23 + 117 k s i r$
Commented by Rasheed.Sindhi last updated on 27/Aug/22

$s i r x = 23 + 117 k {d o e s n}^{'} t s a t i s f y$ $7 x \equiv 2 (m o d 13)$
Commented by cortano1 last updated on 27/Aug/22

$o y e s . i a m t y p o 6 \times 13 = 68$
	Answered by Rasheed.Sindhi last updated on 27/Aug/22

	$4 x \equiv 2 + 2 \times 9 (m o d 9)$ $x \equiv 5 (m o d 9)$ $x = 5 + 9 m$ $7 x \equiv 2 (m o d 13) \Rightarrow 7 (5 + 9 m) \equiv 2 (m o d 13$ $35 + 63 m \equiv 2 (m o d 13)$ $63 m \equiv - 33 (m o d 13)$ $m = 10$ $x = 5 + 9 m = 5 + 9 (10) = 95$ $x = 95 + k \times LCM (9, 13); k \in Z$ $x = 95 + 117 k; k \in Z$
	Commented by Tawa11 last updated on 27/Aug/22

	$Great sir$
	Answered by mr W last updated on 27/Aug/22

	$4 x = 9 k + 2$ $\Rightarrow x = 9 m + 5$ $7 (9 m + 5) = 13 h + 2$ $13 h - 63 m = 33$ $\Rightarrow m = 13 n + 10$ $\Rightarrow x = 9 (13 n + 10) + 5 = 117 n + 95$
	Answered by CElcedricjunior last updated on 28/Aug/22
	${ ((4x≡2[9])),((7x≡2[13])) :}=> { ((28x≡14[9])),((14x≡4[13])) :}=> { ((x≡5[9])),((x≡4[13])) :} =>∃p ;q∈Z/ { ((x=9p+5)),((x=13q+4)) :} or x=x⇔9p−13q=−1 pgcd(9;13)=1 et 1/1 alors cette equation admet de solution soit (−3;−2) une solution particuliee 9p−13q=9(−3)−13(−2) =>9(p+3)=13(q+2)(1) =>9/13(q+2)or pgcd(9;13)=1 =>9/q+2 ⇔∃k∈Z/ q+2=9k (2) =>q=9k−2 (2) dans (1)=>9(p+3)=13(9k) =>p+3=13k=>p=13k−3 de ce qui prece^� de { ((x=9p+5)),((x=13q+4)) :} donc x=9(13k−3)+5=117k−22 ou x=13(9k−2)+4=117k−22 S_Z ={117k−22/k∈Z} S_(IN) ={117k−22/k∈]((22)/(117));→[} .........le ce^� le^� bre cedric junior............$
	${\begin{cases} 4 x \equiv 2 [9] \\ 7 x \equiv 2 [13] \end{cases} => {\begin{cases} 28 x \equiv 14 [9] \\ 14 x \equiv 4 [13] \end{cases} => {\begin{cases} x \equiv 5 [9] \\ x \equiv 4 [13] \end{cases}$ $=> \exists p; q \in Z / {\begin{cases} x = 9 p + 5 \\ x = 13 q + 4 \end{cases}$ $o r x = x \Leftrightarrow 9 p - 13 q = - 1$ $p g c d (9; 13) = 1 e t 1 / 1$ $a l o r s c e t t e e q u a t i o n a d m e t d e s o l u t i o n$ $s o i t (- 3; - 2) u n e s o l u t i o n p a r t i c u l i e e$ $9 p - 13 q = 9 (- 3) - 13 (- 2)$ $=> 9 (p + 3) = 13 (q + 2) (1)$ $=> 9 / 13 (q + 2) o r p g c d (9; 13) = 1$ $=> 9 / q + 2 \Leftrightarrow \exists k \in Z / q + 2 = 9 k (2)$ $=> q = 9 k - 2$ $(2) d a n s (1) => 9 (p + 3) = 13 (9 k)$ $=> p + 3 = 13 k => p = 13 k - 3$ $d e c e q u i p r e c \overset{`}{e d e {\begin{cases} x = 9 p + 5 \\ x = 13 q + 4 \end{cases}}$ $d o n c$ $x = 9 (13 k - 3) + 5 = 117 k - 22$ $o u$ $x = 13 (9 k - 2) + 4 = 117 k - 22$ $S_{Z} = {117 k - 22 / k \in Z}$ $S_{I N} = {117 k - 22 / k \in] \frac{22}{117}; \to [}$ $. . . . . . . . . l e c \overset{´}{e l} \overset{`}{e b r e} c e d r i c j u n i o r . . . . . . . . . . . .$
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