Question Number 121567 by mathocean1 last updated on 09/Nov/20
![Hello how to solve this equation: { ((3x−y≡1[5])),((x+2y≡0[5])) :}](Q121567.png)
$$\mathrm{Hello}\:\mathrm{how}\:\mathrm{to}\:\mathrm{solve}\:\mathrm{this}\: \\ $$$$\mathrm{equation}: \\ $$$$\begin{cases}{\mathrm{3}{x}−{y}\equiv\mathrm{1}\left[\mathrm{5}\right]}\\{{x}+\mathrm{2}{y}\equiv\mathrm{0}\left[\mathrm{5}\right]}\end{cases} \\ $$
Commented by TANMAY PANACEA last updated on 09/Nov/20
![[5] what does [ .] means](Q121573.png)
$$\left[\mathrm{5}\right]\:\:{what}\:{does}\:\left[\:.\right]\:{means} \\ $$
Commented by mathocean1 last updated on 09/Nov/20
$$\mathrm{modulo} \\ $$$$ \\ $$
Answered by mr W last updated on 11/Nov/20
$$\mathrm{3}{x}−{y}=\mathrm{5}{k}+\mathrm{1} \\ $$$$\mathrm{6}{x}−\mathrm{2}{y}=\mathrm{10}{k}+\mathrm{2} \\ $$$${x}+\mathrm{2}{y}=\mathrm{5}{h} \\ $$$$\mathrm{7}{x}=\mathrm{5}\left(\mathrm{2}{k}+{h}\right)+\mathrm{2} \\ $$$$\mathrm{7}\left({x}−\mathrm{1}\right)=\mathrm{5}\left(\mathrm{2}{k}+{h}−\mathrm{1}\right) \\ $$$$\Rightarrow{x}=\mathrm{5}{n}+\mathrm{1} \\ $$$$\Rightarrow{y}=\mathrm{15}{n}−\mathrm{5}{k}+\mathrm{2} \\ $$