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Question Number 134252 by bramlexs22 last updated on 01/Mar/21

 solve  { ((x≡ 2 (mod 3))),((x≡ 5 (mod 7))) :}

$$\:\mathrm{solve}\:\begin{cases}{\mathrm{x}\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{3}\right)}\\{\mathrm{x}\equiv\:\mathrm{5}\:\left(\mathrm{mod}\:\mathrm{7}\right)}\end{cases} \\ $$

Answered by EDWIN88 last updated on 01/Mar/21

 { ((x = 3k+2 ;k∈Z)),((x=7m+5; m∈Z)) :} ⇒ 3k+2 = 7m+5  ⇒ m=((3k−3)/7) ; (k,m)=(8,3),(15,6),(22,9)  ..., (7λ+1, 3λ)  x = 7(3λ)+5 ; x= 21λ+5 , λ∈Z  x = 26, 47, 68, 89, 110, 131,...

$$\begin{cases}{\mathrm{x}\:=\:\mathrm{3k}+\mathrm{2}\:;\mathrm{k}\in\mathbb{Z}}\\{\mathrm{x}=\mathrm{7m}+\mathrm{5};\:\mathrm{m}\in\mathbb{Z}}\end{cases}\:\Rightarrow\:\mathrm{3k}+\mathrm{2}\:=\:\mathrm{7m}+\mathrm{5} \\ $$$$\Rightarrow\:\mathrm{m}=\frac{\mathrm{3k}−\mathrm{3}}{\mathrm{7}}\:;\:\left(\mathrm{k},\mathrm{m}\right)=\left(\mathrm{8},\mathrm{3}\right),\left(\mathrm{15},\mathrm{6}\right),\left(\mathrm{22},\mathrm{9}\right) \\ $$$$...,\:\left(\mathrm{7}\lambda+\mathrm{1},\:\mathrm{3}\lambda\right) \\ $$$$\mathrm{x}\:=\:\mathrm{7}\left(\mathrm{3}\lambda\right)+\mathrm{5}\:;\:\mathrm{x}=\:\mathrm{21}\lambda+\mathrm{5}\:,\:\lambda\in\mathbb{Z} \\ $$$$\mathrm{x}\:=\:\mathrm{26},\:\mathrm{47},\:\mathrm{68},\:\mathrm{89},\:\mathrm{110},\:\mathrm{131},... \\ $$

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