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Question Number 78233 by jagoll last updated on 15/Jan/20
given 5x+22y=18 find for x,y integer
$${given}\:\mathrm{5}{x}+\mathrm{22}{y}=\mathrm{18} \\ $$$${find}\:{for}\:{x},{y}\:{integer} \\ $$
Commented by mr W last updated on 15/Jan/20
x=−22k+8 y=5k−1 with k=any interger
$${x}=−\mathrm{22}{k}+\mathrm{8} \\ $$$${y}=\mathrm{5}{k}−\mathrm{1} \\ $$$${with}\:{k}={any}\:{interger} \\ $$
Commented by jagoll last updated on 15/Jan/20
why −22k+ 8 sir
$${why}\:−\mathrm{22}{k}+\:\mathrm{8}\:{sir} \\ $$
Commented by jagoll last updated on 15/Jan/20
not −22k+18? sir
$${not}\:−\mathrm{22}{k}+\mathrm{18}?\:{sir} \\ $$
Commented by mr W last updated on 15/Jan/20
please read the complete Q19198.
$${please}\:{read}\:{the}\:{complete}\:{Q}\mathrm{19198}. \\ $$
Commented by jagoll last updated on 15/Jan/20
for k = 0 x=8 , y = −1 ⇒5×8+22(−1)=40−22 = 18. right sir
$${for}\:{k}\:=\:\mathrm{0}\: \\ $$$${x}=\mathrm{8}\:,\:{y}\:=\:−\mathrm{1}\:\Rightarrow\mathrm{5}×\mathrm{8}+\mathrm{22}\left(−\mathrm{1}\right)=\mathrm{40}−\mathrm{22}\:=\:\mathrm{18}.\:{right}\:{sir} \\ $$
Commented by jagoll last updated on 15/Jan/20
ok sir. i will read
$${ok}\:{sir}.\:{i}\:{will}\:{read}\: \\ $$
Commented by mathmax by abdo last updated on 15/Jan/20
solution in Z^2 we consider congruence modulo 5 (Z/5Z is corps) (e) ⇒5^− x^− +22^− y^− =18^− ⇒0 +2y^− =3^− =(−2)^− ⇒y^− =(−1)^− ⇒ y=−1+5k (k integr) ⇒5x=18−22(−1+5k) =18+22−110k =40−110k ⇒x =((40−110k)/5) =8−22k ⇒ S ={(8−22k,−1+5k) /k ∈Z}
$${solution}\:{in}\:{Z}^{\mathrm{2}} \:\:{we}\:{consider}\:{congruence}\:{modulo}\:\mathrm{5}\:\:\left({Z}/\mathrm{5}{Z}\:{is}\:{corps}\right) \\ $$$$\left({e}\right)\:\Rightarrow\overset{−} {\mathrm{5}}\overset{−} {{x}}\:+\mathrm{2}\overset{−} {\mathrm{2}}\:\overset{−} {{y}}=\mathrm{1}\overset{−} {\mathrm{8}}\:\Rightarrow\mathrm{0}\:+\mathrm{2}\overset{−} {{y}}=\overset{−} {\mathrm{3}}=\left(−\mathrm{2}\overset{−} {\right)}\:\Rightarrow\overset{−} {{y}}=\left(−\mathrm{1}\overset{−} {\right)}\Rightarrow \\ $$$${y}=−\mathrm{1}+\mathrm{5}{k}\:\:\:\left({k}\:{integr}\right)\:\Rightarrow\mathrm{5}{x}=\mathrm{18}−\mathrm{22}\left(−\mathrm{1}+\mathrm{5}{k}\right)\:=\mathrm{18}+\mathrm{22}−\mathrm{110}{k} \\ $$$$=\mathrm{40}−\mathrm{110}{k}\:\Rightarrow{x}\:=\frac{\mathrm{40}−\mathrm{110}{k}}{\mathrm{5}}\:=\mathrm{8}−\mathrm{22}{k}\:\Rightarrow \\ $$$${S}\:=\left\{\left(\mathrm{8}−\mathrm{22}{k},−\mathrm{1}+\mathrm{5}{k}\right)\:/{k}\:\in{Z}\right\} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 15/Jan/20
5x + 22y = 18 Compare with: ax + by = n a = 5, b = 22, n = 18 d = gcd(5, 22) = 1 ∴ Integer solution of 5x + 22y = 18 x_0 = 8 and y_0 = − 1 Therefore, General solution: x = x_0 − ((bk)/d) and y = y_0 + ((ak)/d) ∴ x = 8 − ((22k)/1) and y = − 1 + ((5k)/1) ∴ x = 8 − 22k and y = − 1 + 5k with k be any integer.
$$\:\:\mathrm{5x}\:+\:\mathrm{22y}\:\:=\:\:\mathrm{18} \\ $$$$\mathrm{Compare}\:\mathrm{with}:\:\:\:\:\mathrm{ax}\:+\:\mathrm{by}\:\:=\:\:\mathrm{n} \\ $$$$\mathrm{a}\:\:=\:\:\mathrm{5},\:\:\:\mathrm{b}\:\:=\:\:\mathrm{22},\:\:\:\:\mathrm{n}\:\:=\:\:\mathrm{18} \\ $$$$\mathrm{d}\:\:=\:\:\mathrm{gcd}\left(\mathrm{5},\:\mathrm{22}\right)\:\:=\:\:\mathrm{1} \\ $$$$\therefore\:\:\:\:\:\mathrm{Integer}\:\mathrm{solution}\:\mathrm{of}\:\:\:\:\mathrm{5x}\:+\:\mathrm{22y}\:\:=\:\:\mathrm{18} \\ $$$$\:\:\mathrm{x}_{\mathrm{0}} \:\:=\:\:\mathrm{8}\:\:\:\:\:\mathrm{and}\:\:\:\mathrm{y}_{\mathrm{0}} \:\:=\:\:−\:\mathrm{1} \\ $$$$\mathrm{Therefore}, \\ $$$$\:\:\:\:\:\:\:\mathrm{General}\:\mathrm{solution}: \\ $$$$\mathrm{x}\:\:\:=\:\:\mathrm{x}_{\mathrm{0}} \:−\:\frac{\mathrm{bk}}{\mathrm{d}}\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\mathrm{y}\:\:=\:\:\mathrm{y}_{\mathrm{0}} \:+\:\frac{\mathrm{ak}}{\mathrm{d}} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:\:=\:\:\mathrm{8}\:−\:\frac{\mathrm{22k}}{\mathrm{1}}\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\mathrm{y}\:\:=\:\:−\:\mathrm{1}\:+\:\frac{\mathrm{5k}}{\mathrm{1}} \\ $$$$\therefore\:\:\:\:\:\mathrm{x}\:\:\:=\:\:\mathrm{8}\:\:−\:\:\mathrm{22k}\:\:\:\:\:\:\:\:\mathrm{and}\:\:\:\:\:\:\:\mathrm{y}\:\:=\:\:−\:\mathrm{1}\:+\:\mathrm{5k} \\ $$$$\mathrm{with}\:\:\mathrm{k}\:\:\mathrm{be}\:\mathrm{any}\:\mathrm{integer}. \\ $$

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