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Please-help-Find-all-the-general-solution-of-6x-8y-5z-101-I-got-x-48-45m-4n-y-48-45m-3n-z-1-2m-Please-

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Question Number 46157 by Tawa1 last updated on 21/Oct/18
Please help. Find all the general solution of 6x + 8y + 5z = 101 . I got: x = − 48 + 45m + 4n y = 48 + 45m − 3n z = 1 − 2m Please help. Am confused.
$$\mathrm{Please}\:\mathrm{help}. \\ $$$$\:\:\:\:\:\mathrm{Find}\:\mathrm{all}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{of}\:\:\:\:\:\:\:\mathrm{6x}\:+\:\mathrm{8y}\:+\:\mathrm{5z}\:=\:\mathrm{101}\:. \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{got}:\:\:\:\:\:\:\:\:\mathrm{x}\:=\:−\:\mathrm{48}\:+\:\mathrm{45m}\:+\:\mathrm{4n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{y}\:=\:\:\:\:\:\mathrm{48}\:+\:\mathrm{45m}\:−\:\mathrm{3n} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\:=\:\:\:\:\:\:\:\mathrm{1}\:−\:\mathrm{2m} \\ $$$$ \\ $$$$\mathrm{Please}\:\mathrm{help}.\:\:\mathrm{Am}\:\mathrm{confused}.\: \\ $$
Answered by MrW3 last updated on 22/Oct/18
6x + 8y + 5z = 101 with x,y,z∈Z since gcd(6,8,5)=1 which divides 101, ⇒solution exists! let 3x+4y=t ...(i) ⇒2t+5z=101 ...(ii) (ii) has a particular solution t_1 =48, z_1 =1 so its general solution is t=48+5m z=1−2m 3x+4y=1 has a particular solution x_1 =−1, y_1 =1 so its general solution is x=−1+4k y=1−3k the general solution for (i) is then x=(−1+4k)(48+5m)=−48−5m+4k(48+5m) y=(1−3k)(48+5m)=48+5m−3k(48+5m) let n=k(48+5m) we get the general solution x=−48−5m+4n y=48+5m−3n z=1−2m check: 6(−48−5m+4n)+8(48+5m−3n)+5(1−2m) =−288−30m+24n+384+40m−24n+5−10m =101 ⇒correct!
$$\mathrm{6x}\:+\:\mathrm{8y}\:+\:\mathrm{5z}\:=\:\mathrm{101}\:{with}\:{x},{y},{z}\in\mathbb{Z} \\ $$$${since}\:{gcd}\left(\mathrm{6},\mathrm{8},\mathrm{5}\right)=\mathrm{1}\:{which}\:{divides}\:\mathrm{101}, \\ $$$$\Rightarrow{solution}\:{exists}! \\ $$$${let}\:\mathrm{3}{x}+\mathrm{4}{y}={t}\:\:\:…\left({i}\right) \\ $$$$\Rightarrow\mathrm{2}{t}+\mathrm{5}{z}=\mathrm{101}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({ii}\right)\:{has}\:{a}\:{particular}\:{solution} \\ $$$${t}_{\mathrm{1}} =\mathrm{48},\:{z}_{\mathrm{1}} =\mathrm{1} \\ $$$${so}\:{its}\:{general}\:{solution}\:{is} \\ $$$${t}=\mathrm{48}+\mathrm{5}{m} \\ $$$${z}=\mathrm{1}−\mathrm{2}{m} \\ $$$$ \\ $$$$\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{1}\:{has}\:{a}\:{particular}\:{solution} \\ $$$${x}_{\mathrm{1}} =−\mathrm{1},\:{y}_{\mathrm{1}} =\mathrm{1} \\ $$$${so}\:{its}\:{general}\:{solution}\:{is} \\ $$$${x}=−\mathrm{1}+\mathrm{4}{k} \\ $$$${y}=\mathrm{1}−\mathrm{3}{k} \\ $$$$ \\ $$$${the}\:{general}\:{solution}\:{for}\:\left({i}\right)\:{is}\:{then} \\ $$$${x}=\left(−\mathrm{1}+\mathrm{4}{k}\right)\left(\mathrm{48}+\mathrm{5}{m}\right)=−\mathrm{48}−\mathrm{5}{m}+\mathrm{4}{k}\left(\mathrm{48}+\mathrm{5}{m}\right) \\ $$$${y}=\left(\mathrm{1}−\mathrm{3}{k}\right)\left(\mathrm{48}+\mathrm{5}{m}\right)=\mathrm{48}+\mathrm{5}{m}−\mathrm{3}{k}\left(\mathrm{48}+\mathrm{5}{m}\right) \\ $$$$ \\ $$$${let}\:{n}={k}\left(\mathrm{48}+\mathrm{5}{m}\right) \\ $$$${we}\:{get}\:{the}\:{general}\:{solution} \\ $$$${x}=−\mathrm{48}−\mathrm{5}{m}+\mathrm{4}{n} \\ $$$${y}=\mathrm{48}+\mathrm{5}{m}−\mathrm{3}{n} \\ $$$${z}=\mathrm{1}−\mathrm{2}{m} \\ $$$$ \\ $$$${check}: \\ $$$$\mathrm{6}\left(−\mathrm{48}−\mathrm{5}{m}+\mathrm{4}{n}\right)+\mathrm{8}\left(\mathrm{48}+\mathrm{5}{m}−\mathrm{3}{n}\right)+\mathrm{5}\left(\mathrm{1}−\mathrm{2}{m}\right) \\ $$$$=−\mathrm{288}−\mathrm{30}{m}+\mathrm{24}{n}+\mathrm{384}+\mathrm{40}{m}−\mathrm{24}{n}+\mathrm{5}−\mathrm{10}{m} \\ $$$$=\mathrm{101}\:\Rightarrow{correct}! \\ $$
Commented by Tawa1 last updated on 21/Oct/18
Ohh, i got it but slight mistake. i followed your method earlier.
$$\mathrm{Ohh},\:\:\mathrm{i}\:\mathrm{got}\:\mathrm{it}\:\mathrm{but}\:\mathrm{slight}\:\mathrm{mistake}.\:\mathrm{i}\:\mathrm{followed}\:\mathrm{your}\:\mathrm{method}\:\mathrm{earlier}. \\ $$
Commented by Tawa1 last updated on 21/Oct/18
Sir, i will always set t = 3x + 4y, later 3x + 4y = 1 ??. Sir why not 3x + 4y = 101 ?
$$\mathrm{Sir},\:\mathrm{i}\:\mathrm{will}\:\mathrm{always}\:\mathrm{set}\:\:\:\:\mathrm{t}\:=\:\mathrm{3x}\:+\:\mathrm{4y},\:\:\:\mathrm{later}\:\:\:\mathrm{3x}\:+\:\mathrm{4y}\:=\:\mathrm{1}\:??. \\ $$$$\mathrm{Sir}\:\mathrm{why}\:\mathrm{not}\:\:\:\:\mathrm{3x}\:+\:\mathrm{4y}\:=\:\mathrm{101}\:? \\ $$
Commented by Tawa1 last updated on 21/Oct/18
And why is the general solution for (i) is x × u i think those are the areas i need to grasp
$$\mathrm{And}\:\mathrm{why}\:\mathrm{is}\:\mathrm{the}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{for}\:\left(\mathrm{i}\right)\:\mathrm{is}\:\:\mathrm{x}\:×\:\mathrm{u} \\ $$$$\mathrm{i}\:\mathrm{think}\:\mathrm{those}\:\mathrm{are}\:\mathrm{the}\:\mathrm{areas}\:\mathrm{i}\:\mathrm{need}\:\mathrm{to}\:\mathrm{grasp} \\ $$
Commented by MrW3 last updated on 23/Oct/18
in the first step you find the solution for 2t+5z=101 which is t=48+5m. in next step you should find the solution for 3x+4y=48+5m. but this is not easy. so you find at first the solution for 3x+4y=1 which is x=−1+4k. the solution for 3x+4y=48+5m is then x=(−1+4k)(48+5m)=−48−5m+4k(48+5m) =−48−5m+4n.
$${in}\:{the}\:{first}\:{step}\:{you}\:{find}\:{the}\:{solution} \\ $$$${for}\:\mathrm{2}{t}+\mathrm{5}{z}=\mathrm{101}\:{which}\:{is}\:{t}=\mathrm{48}+\mathrm{5}{m}. \\ $$$${in}\:{next}\:{step}\:{you}\:{should}\:{find}\:{the} \\ $$$${solution}\:{for}\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{48}+\mathrm{5}{m}.\:{but}\:{this} \\ $$$${is}\:{not}\:{easy}.\:{so}\:{you}\:{find}\:{at}\:{first}\:{the} \\ $$$${solution}\:{for}\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{1}\:{which}\:{is}\:{x}=−\mathrm{1}+\mathrm{4}{k}. \\ $$$${the}\:{solution}\:{for}\:\mathrm{3}{x}+\mathrm{4}{y}=\mathrm{48}+\mathrm{5}{m}\:{is}\:{then} \\ $$$${x}=\left(−\mathrm{1}+\mathrm{4}{k}\right)\left(\mathrm{48}+\mathrm{5}{m}\right)=−\mathrm{48}−\mathrm{5}{m}+\mathrm{4}{k}\left(\mathrm{48}+\mathrm{5}{m}\right) \\ $$$$=−\mathrm{48}−\mathrm{5}{m}+\mathrm{4}{n}. \\ $$
Commented by Tawa1 last updated on 21/Oct/18
Wow, God bless you sir. i understand now
$$\mathrm{Wow},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{understand}\:\mathrm{now} \\ $$

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