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7x-3-mod5-5x-3-mod9-solve-for-x-




Question Number 140779 by 676597498 last updated on 12/May/21
 { ((7x≡3(mod5))),((5x≡3(mod9))) :}  solve for x
$$\begin{cases}{\mathrm{7}{x}\equiv\mathrm{3}\left({mod}\mathrm{5}\right)}\\{\mathrm{5}{x}\equiv\mathrm{3}\left({mod}\mathrm{9}\right)}\end{cases} \\ $$$${solve}\:{for}\:{x} \\ $$
Commented by 676597498 last updated on 12/May/21
pls
$${pls} \\ $$
Answered by mr W last updated on 13/May/21
7x=5p+3 ⇒x=5m+4  5x=9q+3 ⇒x=9n+6  5m+4=9n+6 ⇒m=9k+4  ⇒x=5(9k+4)+4=45k+24  or x≡24 (mod 45)
$$\mathrm{7}{x}=\mathrm{5}{p}+\mathrm{3}\:\Rightarrow{x}=\mathrm{5}{m}+\mathrm{4} \\ $$$$\mathrm{5}{x}=\mathrm{9}{q}+\mathrm{3}\:\Rightarrow{x}=\mathrm{9}{n}+\mathrm{6} \\ $$$$\mathrm{5}{m}+\mathrm{4}=\mathrm{9}{n}+\mathrm{6}\:\Rightarrow{m}=\mathrm{9}{k}+\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{5}\left(\mathrm{9}{k}+\mathrm{4}\right)+\mathrm{4}=\mathrm{45}{k}+\mathrm{24} \\ $$$${or}\:{x}\equiv\mathrm{24}\:\left({mod}\:\mathrm{45}\right) \\ $$

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