Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 163770 by mnjuly1970 last updated on 10/Jan/22

solve : x,y ∈ N 3x + 5y = 20 −−−−−−−

$$ \\ $$$$\:\:\:\:\:{solve}\::\:\:\:\:{x},{y}\:\in\:\mathbb{N}\: \\ $$$$\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\mathrm{3}{x}\:+\:\mathrm{5}{y}\:=\:\mathrm{20}\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−−−−−−− \\ $$

Answered by MJS_new last updated on 10/Jan/22

y=4−(3/5)x ⇒ (x=0∧y=4)∨(x=5∧y=1)

$${y}=\mathrm{4}−\frac{\mathrm{3}}{\mathrm{5}}{x}\:\Rightarrow\:\left({x}=\mathrm{0}\wedge{y}=\mathrm{4}\right)\vee\left({x}=\mathrm{5}\wedge{y}=\mathrm{1}\right) \\ $$

Commented by MJS_new last updated on 10/Jan/22

yes, thank you!

$$\mathrm{yes},\:\mathrm{thank}\:\mathrm{you}! \\ $$

Commented by MathsFan last updated on 10/Jan/22

sir, which method did you use please

$${sir},\:{which}\:{method}\:{did}\:{you}\:{use}\:{please} \\ $$

Commented by MJS_new last updated on 10/Jan/22

it′s obvious for x, y ∈N, no special method needed. 3x+5y=20 ⇔ y=4−(3/5)x ⇒ ⇒ x=5n ⇒ y=4−3n ⇒ n=0∨n=1

$$\mathrm{it}'\mathrm{s}\:\mathrm{obvious}\:\mathrm{for}\:{x},\:{y}\:\in\mathbb{N},\:\mathrm{no}\:\mathrm{special}\:\mathrm{method} \\ $$$$\mathrm{needed}. \\ $$$$\mathrm{3}{x}+\mathrm{5}{y}=\mathrm{20}\:\Leftrightarrow\:{y}=\mathrm{4}−\frac{\mathrm{3}}{\mathrm{5}}{x}\:\Rightarrow \\ $$$$\Rightarrow\:{x}=\mathrm{5}{n}\:\Rightarrow\:{y}=\mathrm{4}−\mathrm{3}{n}\:\Rightarrow\:{n}=\mathrm{0}\vee{n}=\mathrm{1} \\ $$

Answered by Mathspace last updated on 10/Jan/22

we consider congruence modulo3 e ⇒3^. x^. +5^. y^. =(20)^. ⇒ 0+2^. y^. =2^. ⇒y^. =1^. ⇒y=3k+1 3x=20−5y =20−5(3k+1) =15−15k (k∈Z) rememeber thst Z/3Z est un corps

$${we}\:{consider}\:{congruence}\:{modulo}\mathrm{3} \\ $$$${e}\:\Rightarrow\overset{.} {\mathrm{3}}\overset{.} {{x}}\:+\overset{.} {\mathrm{5}}\overset{.} {{y}}\:=\left(\mathrm{20}\overset{.} {\right)}\:\Rightarrow \\ $$$$\mathrm{0}+\overset{.} {\mathrm{2}}\overset{.} {{y}}=\overset{.} {\mathrm{2}}\:\:\Rightarrow\overset{.} {{y}}=\overset{.} {\mathrm{1}}\:\Rightarrow{y}=\mathrm{3}{k}+\mathrm{1} \\ $$$$\mathrm{3}{x}=\mathrm{20}−\mathrm{5}{y}\:=\mathrm{20}−\mathrm{5}\left(\mathrm{3}{k}+\mathrm{1}\right) \\ $$$$=\mathrm{15}−\mathrm{15}{k}\:\:\:\left({k}\in{Z}\right) \\ $$$${rememeber}\:{thst}\:{Z}/\mathrm{3}{Z}\:\:{est}\:{un}\:{corps} \\ $$

Commented by Mathspace last updated on 10/Jan/22

3x=15−15k ⇒x=5−5k the set of solution is {(5−5k,1+3k)} another way on a (5,1) is special solution because 3×5+5.1=20 we have 3x+5y=20 and 3.5+5.1=20 ⇒ 3(x−1)+5(y−1)=0 ⇒ 3(x−1)=5(1−y) ⇒ 3divide5(1−y) but Δ(3,5)=1 bizout ⇒3/1−y ⇒1−y=3k ⇒ y=−3k+1 5/3(x−1)⇒5/x−1 ⇒x=5k+1

$$\mathrm{3}{x}=\mathrm{15}−\mathrm{15}{k}\:\Rightarrow{x}=\mathrm{5}−\mathrm{5}{k} \\ $$$${the}\:{set}\:{of}\:{solution}\:{is} \\ $$$$\left\{\left(\mathrm{5}−\mathrm{5}{k},\mathrm{1}+\mathrm{3}{k}\right)\right\} \\ $$$${another}\:{way} \\ $$$${on}\:{a}\:\left(\mathrm{5},\mathrm{1}\right)\:{is}\:{special}\:{solution} \\ $$$${because}\:\mathrm{3}×\mathrm{5}+\mathrm{5}.\mathrm{1}=\mathrm{20}\:{we}\:{have} \\ $$$$\mathrm{3}{x}+\mathrm{5}{y}=\mathrm{20}\:{and}\:\mathrm{3}.\mathrm{5}+\mathrm{5}.\mathrm{1}=\mathrm{20}\:\Rightarrow \\ $$$$\mathrm{3}\left({x}−\mathrm{1}\right)+\mathrm{5}\left({y}−\mathrm{1}\right)=\mathrm{0}\:\Rightarrow \\ $$$$\mathrm{3}\left({x}−\mathrm{1}\right)=\mathrm{5}\left(\mathrm{1}−{y}\right)\:\Rightarrow \\ $$$$\mathrm{3}{divide}\mathrm{5}\left(\mathrm{1}−{y}\right)\:{but}\:\Delta\left(\mathrm{3},\mathrm{5}\right)=\mathrm{1} \\ $$$${bizout}\:\Rightarrow\mathrm{3}/\mathrm{1}−{y}\:\Rightarrow\mathrm{1}−{y}=\mathrm{3}{k}\:\Rightarrow \\ $$$${y}=−\mathrm{3}{k}+\mathrm{1} \\ $$$$\mathrm{5}/\mathrm{3}\left({x}−\mathrm{1}\right)\Rightarrow\mathrm{5}/{x}−\mathrm{1}\:\Rightarrow{x}=\mathrm{5}{k}+\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com