Question Number 175320 by cortano1 last updated on 27/Aug/22
$$\:\:\begin{cases}{\mathrm{4}{x}=\mathrm{2}\left({mod}\:\mathrm{9}\right)}\\{\mathrm{7}{x}=\mathrm{2}\:\left({mod}\:\mathrm{13}\right)}\end{cases} \\ $$
Commented by cortano1 last updated on 27/Aug/22
$${i}\:{got}\:{x}=\mathrm{23}\:+\mathrm{117}{k}\:{sir} \\ $$
Commented by Rasheed.Sindhi last updated on 27/Aug/22
$${sir}\:{x}=\mathrm{23}+\mathrm{117}{k}\:\:{doesn}'{t}\:{satisfy} \\ $$$$\mathrm{7}{x}\equiv\mathrm{2}\left({mod}\:\mathrm{13}\right) \\ $$
Commented by cortano1 last updated on 27/Aug/22
$${o}\:{yes}.\:{i}\:{am}\:{typo}\:\mathrm{6}×\mathrm{13}=\mathrm{68}\: \\ $$
Answered by Rasheed.Sindhi last updated on 27/Aug/22
$$\mathrm{4}{x}\equiv\mathrm{2}+\mathrm{2}×\mathrm{9}\left({mod}\:\mathrm{9}\right) \\ $$$${x}\equiv\mathrm{5}\left({mod}\:\mathrm{9}\right) \\ $$$${x}=\mathrm{5}+\mathrm{9}{m} \\ $$$$\mathrm{7}{x}\equiv\mathrm{2}\left({mod}\:\mathrm{13}\right)\Rightarrow\mathrm{7}\left(\mathrm{5}+\mathrm{9}{m}\right)\equiv\mathrm{2}\left({mod}\:\mathrm{13}\right. \\ $$$$\mathrm{35}+\mathrm{63}{m}\equiv\mathrm{2}\left({mod}\:\mathrm{13}\right) \\ $$$$\mathrm{63}{m}\equiv−\mathrm{33}\left({mod}\:\mathrm{13}\right) \\ $$$${m}=\mathrm{10} \\ $$$${x}=\mathrm{5}+\mathrm{9}{m}=\mathrm{5}+\mathrm{9}\left(\mathrm{10}\right)=\mathrm{95} \\ $$$${x}=\mathrm{95}+{k}×\mathrm{LCM}\left(\mathrm{9},\mathrm{13}\right)\:;{k}\in\mathbb{Z} \\ $$$${x}=\mathrm{95}+\mathrm{117}{k}\:;{k}\in\mathbb{Z} \\ $$
Commented by Tawa11 last updated on 27/Aug/22
$$\mathrm{Great}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 27/Aug/22
$$\mathrm{4}{x}=\mathrm{9}{k}+\mathrm{2} \\ $$$$\Rightarrow{x}=\mathrm{9}{m}+\mathrm{5} \\ $$$$\mathrm{7}\left(\mathrm{9}{m}+\mathrm{5}\right)=\mathrm{13}{h}+\mathrm{2} \\ $$$$\mathrm{13}{h}−\mathrm{63}{m}=\mathrm{33} \\ $$$$\Rightarrow{m}=\mathrm{13}{n}+\mathrm{10} \\ $$$$\Rightarrow{x}=\mathrm{9}\left(\mathrm{13}{n}+\mathrm{10}\right)+\mathrm{5}=\mathrm{117}{n}+\mathrm{95} \\ $$
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............](https://www.tinkutara.com/question/Q175367.png)
$$\begin{cases}{\mathrm{4}\boldsymbol{{x}}\equiv\mathrm{2}\left[\mathrm{9}\right]}\\{\mathrm{7}\boldsymbol{{x}}\equiv\mathrm{2}\left[\mathrm{13}\right]}\end{cases}=>\begin{cases}{\mathrm{28}\boldsymbol{{x}}\equiv\mathrm{14}\left[\mathrm{9}\right]}\\{\mathrm{14}\boldsymbol{{x}}\equiv\mathrm{4}\left[\mathrm{13}\right]}\end{cases}=>\begin{cases}{\boldsymbol{{x}}\equiv\mathrm{5}\left[\mathrm{9}\right]}\\{\boldsymbol{{x}}\equiv\mathrm{4}\left[\mathrm{13}\right]}\end{cases} \\ $$$$=>\exists\boldsymbol{{p}}\:;\boldsymbol{{q}}\in\mathbb{Z}/\begin{cases}{\boldsymbol{{x}}=\mathrm{9}\boldsymbol{{p}}+\mathrm{5}}\\{\boldsymbol{{x}}=\mathrm{13}\boldsymbol{{q}}+\mathrm{4}}\end{cases} \\ $$$$\boldsymbol{{or}}\:\boldsymbol{{x}}=\boldsymbol{{x}}\Leftrightarrow\mathrm{9}\boldsymbol{{p}}−\mathrm{13}\boldsymbol{{q}}=−\mathrm{1} \\ $$$$\boldsymbol{{pgcd}}\left(\mathrm{9};\mathrm{13}\right)=\mathrm{1}\:{et}\:\mathrm{1}/\mathrm{1} \\ $$$$\:\boldsymbol{{alors}}\:\boldsymbol{{cette}}\:\boldsymbol{{equation}}\:\boldsymbol{{admet}}\:\boldsymbol{{de}}\:\boldsymbol{{solution}} \\ $$$$\boldsymbol{{soit}}\:\left(−\mathrm{3};−\mathrm{2}\right)\:\boldsymbol{{une}}\:\boldsymbol{{solution}}\:\boldsymbol{{particuliee}} \\ $$$$\mathrm{9}\boldsymbol{{p}}−\mathrm{13}\boldsymbol{{q}}=\mathrm{9}\left(−\mathrm{3}\right)−\mathrm{13}\left(−\mathrm{2}\right) \\ $$$$=>\mathrm{9}\left(\boldsymbol{{p}}+\mathrm{3}\right)=\mathrm{13}\left(\boldsymbol{{q}}+\mathrm{2}\right)\left(\mathrm{1}\right) \\ $$$$=>\mathrm{9}/\mathrm{13}\left(\boldsymbol{{q}}+\mathrm{2}\right)\boldsymbol{{or}}\:\boldsymbol{{pgcd}}\left(\mathrm{9};\mathrm{13}\right)=\mathrm{1} \\ $$$$=>\mathrm{9}/\boldsymbol{{q}}+\mathrm{2}\:\Leftrightarrow\exists\boldsymbol{{k}}\in\mathbb{Z}/\:\boldsymbol{{q}}+\mathrm{2}=\mathrm{9}\boldsymbol{{k}}\:\:\left(\mathrm{2}\right) \\ $$$$=>\boldsymbol{{q}}=\mathrm{9}\boldsymbol{{k}}−\mathrm{2} \\ $$$$\left(\mathrm{2}\right)\:\boldsymbol{{dans}}\:\left(\mathrm{1}\right)=>\mathrm{9}\left(\boldsymbol{{p}}+\mathrm{3}\right)=\mathrm{13}\left(\mathrm{9}\boldsymbol{{k}}\right) \\ $$$$=>\boldsymbol{{p}}+\mathrm{3}=\mathrm{13}\boldsymbol{{k}}=>\boldsymbol{{p}}=\mathrm{13}\boldsymbol{{k}}−\mathrm{3} \\ $$$$\boldsymbol{{de}}\:\boldsymbol{{ce}}\:\boldsymbol{{qui}}\:\boldsymbol{{prec}}\grave {\boldsymbol{{e}de\begin{cases}{\boldsymbol{{x}}=\mathrm{9}\boldsymbol{{p}}+\mathrm{5}}\\{\boldsymbol{{x}}=\mathrm{13}\boldsymbol{{q}}+\mathrm{4}}\end{cases}}}\: \\ $$$$\boldsymbol{{donc}}\: \\ $$$$\boldsymbol{{x}}=\mathrm{9}\left(\mathrm{13}\boldsymbol{{k}}−\mathrm{3}\right)+\mathrm{5}=\mathrm{117}\boldsymbol{{k}}−\mathrm{22} \\ $$$$\boldsymbol{{ou}} \\ $$$$\boldsymbol{{x}}=\mathrm{13}\left(\mathrm{9}\boldsymbol{{k}}−\mathrm{2}\right)+\mathrm{4}=\mathrm{117}\boldsymbol{{k}}−\mathrm{22} \\ $$$$\boldsymbol{{S}}_{\mathbb{Z}} =\left\{\mathrm{117}\boldsymbol{{k}}−\mathrm{22}/\boldsymbol{{k}}\in\mathbb{Z}\right\} \\ $$$${S}_{\boldsymbol{{IN}}} =\left\{\mathrm{117}\boldsymbol{{k}}−\mathrm{22}/\boldsymbol{{k}}\in\right]\frac{\mathrm{22}}{\mathrm{117}};\rightarrow\left[\right\} \\ $$$$\: \\ $$$$\:………{le}\:{c}\acute {{e}l}\grave {{e}bre}\:{cedric}\:{junior}………… \\ $$$$ \\ $$