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Question Number 64086 by Rio Michael last updated on 12/Jul/19

if 3x + 5y = 1 use Bezout′s identity to find the value of x and y

if3x+5y=1useBezout′sidentitytofindthevalueofxandy

Commented by turbo msup by abdo last updated on 13/Jul/19

let consider congruence [3] (Z/3Z) (e) ⇒3^− x^− +5^− y^− =1^− ⇒ 0+2^− y^− =1^− =−2^− ⇒y^− =−1^− ⇒ y=−1+3k kintegr ⇒ x=((1−5y)/3) =((1+5−15k)/3) =2−5k ⇒ the solution are (2−5k,−1+3k) kfromZ .

letconsidercongruence[3](Z/3Z)(e)⇒3−x−+5−y−=1−⇒0+2−y−=1−=−2−⇒y−=−1−⇒y=−1+3kkintegr⇒x=1−5y3=1+5−15k3=2−5k⇒thesolutionare(2−5k,−1+3k)kfromZ.

Commented by turbo msup by abdo last updated on 13/Jul/19

another way by particular solution (2,−1) is a osrticular solution we have 3x+5y=1 and 3×2+5×(−1)=1 ⇒ 3(x−2)+5(y+1)=0 ⇒ 3(x−2)=−5(y+1) ⇒ 5 divide 3(x−2) but Δ(3,5)=1⇒ 5 divide x−2 ⇒x=5k+2 we have y=((1−3x)/5) =((1−15k−6)/5) =−1−3k so (5k+2,−1−3k) are solution for this equation.

anotherwaybyparticularsolution(2,−1)isaosrticularsolutionwehave3x+5y=1and3×2+5×(−1)=1⇒3(x−2)+5(y+1)=0⇒3(x−2)=−5(y+1)⇒5divide3(x−2)butΔ(3,5)=1⇒5dividex−2⇒x=5k+2wehavey=1−3x5=1−15k−65=−1−3kso(5k+2,−1−3k)aresolutionforthisequation.

Commented by Rio Michael last updated on 13/Jul/19

i thought we are to get constants as answers. we use gcd(3,5)=1 right? since they are relativey prime

ithoughtwearetogetconstantsasanswers.weusegcd(3,5)=1right?sincetheyarerelativeyprime

Commented by mr W last updated on 13/Jul/19

see also Q46592

seealsoQ46592

Commented by Rio Michael last updated on 13/Jul/19

yes i saw that,but must we always put it in a general solution since gcd(3,5)=1 then we reverse right x=−2 and y=1

yesisawthat,butmustwealwaysputitinageneralsolutionsincegcd(3,5)=1thenwereverserightx=−2andy=1

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