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Question Number 110281 by 675480065 last updated on 28/Aug/20

Answered by Rio Michael last updated on 28/Aug/20

3x ≡ 1 (mod 5) (i) 5x ≡ 2 (mod 7) (ii) from (i) x ≡ 2 (mod 5) (iii) from (ii) x ≡ 6 (mod 7) (iv) from (iii) x = 5k + 2 , k ∈ N (v) into (iv) ⇒ 5k + 2 ≡ 6 (mod 7) ⇒ 5k ≡ 4 (mod 7 ) ⇔ k ≡ 5( mod 7) ⇒ k = 7t + 5 , t ∈ N hence x = 5(7t + 5) + 2 = 35t + 27 x = 35t + 27 ,

$$\:\mathrm{3}{x}\:\equiv\:\mathrm{1}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\:\:\left({i}\right) \\ $$$$\:\mathrm{5}{x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\:\:\:\:\left({ii}\right) \\ $$$$\mathrm{from}\:\left({i}\right)\:\:{x}\:\equiv\:\mathrm{2}\:\left(\mathrm{mod}\:\mathrm{5}\right)\:\:\left({iii}\right) \\ $$$$\mathrm{from}\:\left({ii}\right)\:{x}\:\equiv\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{7}\right)\:\:\left({iv}\right) \\ $$$$\:\mathrm{from}\:\left({iii}\right)\:\:{x}\:=\:\mathrm{5}{k}\:+\:\mathrm{2}\:,\:\:{k}\:\in\:\mathbb{N}\:\left({v}\right) \\ $$$$\mathrm{into}\:\left({iv}\right)\:\Rightarrow\:\mathrm{5}{k}\:+\:\mathrm{2}\:\equiv\:\mathrm{6}\:\left(\mathrm{mod}\:\mathrm{7}\right)\: \\ $$$$\:\Rightarrow\:\mathrm{5}{k}\:\equiv\:\mathrm{4}\:\left(\mathrm{mod}\:\mathrm{7}\:\right)\:\Leftrightarrow\:\:{k}\:\equiv\:\mathrm{5}\left(\:\mathrm{mod}\:\mathrm{7}\right)\:\Rightarrow\:{k}\:=\:\mathrm{7}{t}\:+\:\mathrm{5}\:,\:{t}\:\in\:\mathbb{N} \\ $$$$\mathrm{hence}\:{x}\:=\:\mathrm{5}\left(\mathrm{7}{t}\:+\:\mathrm{5}\right)\:+\:\mathrm{2}\:=\:\mathrm{35}{t}\:+\:\mathrm{27} \\ $$$${x}\:=\:\mathrm{35}{t}\:+\:\mathrm{27}\:, \\ $$$$\:\:\: \\ $$$$\: \\ $$

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