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Question Number 46592 by Tawa1 last updated on 29/Oct/18
Find a solution to: 7x + 5y + 15z + 12w = 149
$$\mathrm{Find}\:\mathrm{a}\:\mathrm{solution}\:\mathrm{to}:\:\:\:\mathrm{7x}\:+\:\mathrm{5y}\:+\:\mathrm{15z}\:+\:\mathrm{12w}\:=\:\mathrm{149} \\ $$
Commented by Tawa1 last updated on 29/Oct/18
No general solution sir
$$\mathrm{No}\:\mathrm{general}\:\mathrm{solution}\:\mathrm{sir} \\ $$
Answered by MrW3 last updated on 30/Oct/18
7x+5y+15z+12w=149 with x,y,z,w∈Z gcd(7,5,15,12)=1 which divides 149, ⇒solution exists. the original eqn. can be formed into: 7x+5(y+3z)+12w=149 or y+3z=u ...(iii) 7x+5u=v ...(ii) v+12w=149 ...(i) eqn. (i): v=5 and w=12 is a particular solution, ⇒v=5+12m ⇒w=12−m eqn. (ii): 7x+5u=1 has a particular solution x=3, u=−4 its general solution is x=3+5i, u=−4−7i the general solution for 7x+5u=v is then x=(3+5i)v=(3+5i)(5+12m)=15+36m+5i(5+12m)=15+36m+5n u=(−4−7i)v=(−4−7i)(5+12m)=−20−48m−7i(5+12m)=−20−48m−7n eqn. (i): y+3z=1 has a particular solution y=−2, z=1 its general solution is y=−2+3j, z=1−j the general solution for y+3z=u is then y=(−2+3j)u=(−2+3j)(−20−48m−7n)=40+96m+14n+3j(−20−48m−7n)=40+96m+14n+3k z=(1−j)u=(1−j)(−20−48m−7n)=−20−48m−7n−j(−20−48m−7n)=−20−48m−7n−k ⇒ { ((x=15+36m+5n)),((y=40+96m+14n+3k)),((z=−20−48m−7n−k)),((w=12−m)) :} with m,n,k ∈ Z
$$\mathrm{7}{x}+\mathrm{5}{y}+\mathrm{15}{z}+\mathrm{12}{w}=\mathrm{149} \\ $$$${with}\:{x},{y},{z},{w}\in\mathbb{Z} \\ $$$${gcd}\left(\mathrm{7},\mathrm{5},\mathrm{15},\mathrm{12}\right)=\mathrm{1}\:{which}\:{divides}\:\mathrm{149}, \\ $$$$\Rightarrow{solution}\:{exists}. \\ $$$$ \\ $$$${the}\:{original}\:{eqn}.\:{can}\:{be}\:{formed}\:{into}: \\ $$$$\mathrm{7}{x}+\mathrm{5}\left({y}+\mathrm{3}{z}\right)+\mathrm{12}{w}=\mathrm{149} \\ $$$${or} \\ $$$${y}+\mathrm{3}{z}={u}\:\:\:\:…\left({iii}\right) \\ $$$$\mathrm{7}{x}+\mathrm{5}{u}={v}\:\:\:\:\:…\left({ii}\right) \\ $$$${v}+\mathrm{12}{w}=\mathrm{149}\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${eqn}.\:\left({i}\right): \\ $$$${v}=\mathrm{5}\:{and}\:{w}=\mathrm{12}\:{is}\:{a}\:{particular}\:{solution}, \\ $$$$\Rightarrow{v}=\mathrm{5}+\mathrm{12}{m} \\ $$$$\Rightarrow{w}=\mathrm{12}−{m} \\ $$$$ \\ $$$${eqn}.\:\left({ii}\right): \\ $$$$\mathrm{7}{x}+\mathrm{5}{u}=\mathrm{1}\:{has}\:{a}\:{particular}\:{solution} \\ $$$${x}=\mathrm{3},\:{u}=−\mathrm{4} \\ $$$${its}\:{general}\:{solution}\:{is} \\ $$$${x}=\mathrm{3}+\mathrm{5}{i},\:{u}=−\mathrm{4}−\mathrm{7}{i} \\ $$$${the}\:{general}\:{solution}\:{for} \\ $$$$\mathrm{7}{x}+\mathrm{5}{u}={v}\:{is}\:{then} \\ $$$${x}=\left(\mathrm{3}+\mathrm{5}{i}\right){v}=\left(\mathrm{3}+\mathrm{5}{i}\right)\left(\mathrm{5}+\mathrm{12}{m}\right)=\mathrm{15}+\mathrm{36}{m}+\mathrm{5}{i}\left(\mathrm{5}+\mathrm{12}{m}\right)=\mathrm{15}+\mathrm{36}{m}+\mathrm{5}{n} \\ $$$${u}=\left(−\mathrm{4}−\mathrm{7}{i}\right){v}=\left(−\mathrm{4}−\mathrm{7}{i}\right)\left(\mathrm{5}+\mathrm{12}{m}\right)=−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{i}\left(\mathrm{5}+\mathrm{12}{m}\right)=−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n} \\ $$$$ \\ $$$${eqn}.\:\left({i}\right): \\ $$$${y}+\mathrm{3}{z}=\mathrm{1}\:{has}\:{a}\:{particular}\:{solution} \\ $$$${y}=−\mathrm{2},\:{z}=\mathrm{1} \\ $$$${its}\:{general}\:{solution}\:{is} \\ $$$${y}=−\mathrm{2}+\mathrm{3}{j},\:{z}=\mathrm{1}−{j} \\ $$$${the}\:{general}\:{solution}\:{for} \\ $$$${y}+\mathrm{3}{z}={u}\:{is}\:{then} \\ $$$${y}=\left(−\mathrm{2}+\mathrm{3}{j}\right){u}=\left(−\mathrm{2}+\mathrm{3}{j}\right)\left(−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}\right)=\mathrm{40}+\mathrm{96}{m}+\mathrm{14}{n}+\mathrm{3}{j}\left(−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}\right)=\mathrm{40}+\mathrm{96}{m}+\mathrm{14}{n}+\mathrm{3}{k} \\ $$$${z}=\left(\mathrm{1}−{j}\right){u}=\left(\mathrm{1}−{j}\right)\left(−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}\right)=−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}−{j}\left(−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}\right)=−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}−{k} \\ $$$$ \\ $$$$\Rightarrow\begin{cases}{{x}=\mathrm{15}+\mathrm{36}{m}+\mathrm{5}{n}}\\{{y}=\mathrm{40}+\mathrm{96}{m}+\mathrm{14}{n}+\mathrm{3}{k}}\\{{z}=−\mathrm{20}−\mathrm{48}{m}−\mathrm{7}{n}−{k}}\\{{w}=\mathrm{12}−{m}}\end{cases} \\ $$$${with}\:{m},{n},{k}\:\in\:\mathbb{Z} \\ $$
Commented by Tawa1 last updated on 29/Oct/18
God bless you sir.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by Tawa1 last updated on 30/Oct/18
Please sir, i don′t get: x = (3 + 5i) how you got it (3 + 5i) u = (− 4 − 7i) how you got it (− 4 − 7i) y = (− 2 + 3j) how you got it (− 2 + 3j) z = (1 − j) how you got it (1 − j) I found those steps difficult to understand sir. I will be happy if you can break it down sir. Thanks for every time. God bless you sir.
$$\mathrm{Please}\:\mathrm{sir},\:\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}: \\ $$$$\:\:\:\:\mathrm{x}\:=\:\left(\mathrm{3}\:+\:\mathrm{5i}\right)\:\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{it}\:\:\:\:\:\:\:\:\left(\mathrm{3}\:+\:\mathrm{5i}\right) \\ $$$$\:\:\:\:\mathrm{u}\:=\:\left(−\:\mathrm{4}\:−\:\mathrm{7i}\right)\:\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{it}\:\:\:\:\:\:\:\:\left(−\:\mathrm{4}\:−\:\mathrm{7i}\right) \\ $$$$\:\:\:\:\mathrm{y}\:=\:\left(−\:\mathrm{2}\:+\:\mathrm{3j}\right)\:\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{it}\:\:\:\:\:\:\:\:\left(−\:\mathrm{2}\:+\:\mathrm{3j}\right) \\ $$$$\:\:\:\:\mathrm{z}\:=\:\left(\mathrm{1}\:−\:\mathrm{j}\right)\:\:\mathrm{how}\:\mathrm{you}\:\mathrm{got}\:\mathrm{it}\:\:\:\:\:\:\:\:\left(\mathrm{1}\:−\:\mathrm{j}\right) \\ $$$$ \\ $$$$\mathrm{I}\:\mathrm{found}\:\mathrm{those}\:\mathrm{steps}\:\mathrm{difficult}\:\mathrm{to}\:\mathrm{understand}\:\mathrm{sir}. \\ $$$$\:\:\:\:\mathrm{I}\:\mathrm{will}\:\mathrm{be}\:\mathrm{happy}\:\mathrm{if}\:\mathrm{you}\:\mathrm{can}\:\mathrm{break}\:\mathrm{it}\:\mathrm{down}\:\mathrm{sir}.\: \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{every}\:\mathrm{time}.\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Commented by MrW3 last updated on 30/Oct/18
what I can explain is explained in Q46157. besides I have put more steps into my working above. Or do you mean you just have generally problem to find a particular solution for each of the equations?
$${what}\:{I}\:{can}\:{explain}\:{is}\:{explained}\:{in}\:{Q}\mathrm{46157}. \\ $$$${besides}\:{I}\:{have}\:{put}\:{more}\:{steps}\:{into} \\ $$$${my}\:{working}\:{above}. \\ $$$${Or}\:{do}\:{you}\:{mean}\:{you}\:{just}\:{have}\:{generally} \\ $$$${problem}\:{to}\:{find}\:{a}\:{particular}\:{solution} \\ $$$${for}\:{each}\:{of}\:{the}\:{equations}? \\ $$
Commented by Tawa1 last updated on 30/Oct/18
Wow, i understand better now with the steps you add in your solution, God bless you sir
$$\mathrm{Wow},\:\mathrm{i}\:\mathrm{understand}\:\mathrm{better}\:\mathrm{now}\:\mathrm{with}\:\mathrm{the}\:\mathrm{steps}\:\mathrm{you}\:\mathrm{add}\:\mathrm{in}\:\mathrm{your}\: \\ $$$$\mathrm{solution},\:\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 30/Oct/18
from equation (ii) sir. you type mistake but corrected it in next line x = 3 + 5i, u = − 4 − 7i. But sir, why from u = (− 4 − 7i)v = ....... = − 20 − 48m − 7i(5 + 12m) Now, − 7i(5 + 12m) becomes 7n. Reason sir.
$$\mathrm{from}\:\mathrm{equation}\:\left(\mathrm{ii}\right)\:\mathrm{sir}.\:\mathrm{you}\:\mathrm{type}\:\mathrm{mistake}\:\mathrm{but}\:\mathrm{corrected}\:\mathrm{it}\:\mathrm{in}\:\mathrm{next}\:\mathrm{line} \\ $$$$\mathrm{x}\:=\:\mathrm{3}\:+\:\mathrm{5i},\:\mathrm{u}\:=\:−\:\mathrm{4}\:−\:\mathrm{7i}. \\ $$$$ \\ $$$$\mathrm{But}\:\mathrm{sir},\:\:\mathrm{why}\:\mathrm{from}\:\:\mathrm{u}\:=\:\left(−\:\mathrm{4}\:−\:\mathrm{7i}\right)\mathrm{v}\:\:=\:\:…….\:=\:\:−\:\mathrm{20}\:−\:\mathrm{48m}\:−\:\mathrm{7i}\left(\mathrm{5}\:+\:\mathrm{12m}\right) \\ $$$$ \\ $$$$\mathrm{Now},\:\:\:\:\:−\:\mathrm{7i}\left(\mathrm{5}\:+\:\mathrm{12m}\right)\:\:\mathrm{becomes}\:\:\:\:\mathrm{7n}.\:\:\:\:\mathrm{Reason}\:\mathrm{sir}. \\ $$
Commented by MrW3 last updated on 30/Oct/18
i(5 + 12m) is an integer which can be notated as n or what ever you like.
$$\mathrm{i}\left(\mathrm{5}\:+\:\mathrm{12m}\right)\:{is}\:{an}\:{integer}\:{which}\:{can}\:{be} \\ $$$${notated}\:{as}\:{n}\:{or}\:{what}\:{ever}\:{you}\:{like}. \\ $$
Commented by Tawa1 last updated on 30/Oct/18
God bless you sir. I understand everything now.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{understand}\:\mathrm{everything}\:\mathrm{now}.\:\: \\ $$
Commented by MrW3 last updated on 31/Oct/18
nice to know this sir! just as an exercise please try 133x+65y−211z+47w=93
$${nice}\:{to}\:{know}\:{this}\:{sir}! \\ $$$${just}\:{as}\:{an}\:{exercise}\:{please}\:{try} \\ $$$$\mathrm{133}{x}+\mathrm{65}{y}−\mathrm{211}{z}+\mathrm{47}{w}=\mathrm{93} \\ $$
Commented by MrW3 last updated on 31/Oct/18
I have put some large numbers such that a particular solution can not be found obviously.
$${I}\:{have}\:{put}\:{some}\:{large}\:{numbers}\:{such}\:{that} \\ $$$${a}\:{particular}\:{solution}\:{can}\:{not}\:{be}\:{found} \\ $$$${obviously}. \\ $$
Commented by Tawa1 last updated on 31/Oct/18
133x + 65y + 47w − 211z = 93 133x + 65y = u ...... i 47w − 211z = v ..... ii u + v = 93 ....... iii From (iii) u + v = 93 u = 1 + m, v = 92 − m from (ii) 47w − 211z = v ..... ii w = 828 − 9m − 211n z = 184 − 2m − 47n from (i) 133x + 65y = u ...... i x = 22 + 22m + 65k y = − 45 − 45m − 133k Now x = 22 + 22m + 65k y = − 45 − 45m − 133k z = 184 − 2m − 47n w = 828 − 9m − 211n
$$\mathrm{133x}\:+\:\mathrm{65y}\:+\:\mathrm{47w}\:−\:\mathrm{211z}\:=\:\mathrm{93} \\ $$$$\mathrm{133x}\:+\:\mathrm{65y}\:=\:\mathrm{u}\:\:\:\:\:\:\:……\:\mathrm{i} \\ $$$$\mathrm{47w}\:−\:\mathrm{211z}\:=\:\mathrm{v}\:\:\:\:\:…..\:\mathrm{ii} \\ $$$$\mathrm{u}\:+\:\mathrm{v}\:=\:\mathrm{93}\:\:\:\:\:\:\:…….\:\mathrm{iii} \\ $$$$ \\ $$$$\mathrm{From}\:\left(\mathrm{iii}\right) \\ $$$$\mathrm{u}\:+\:\mathrm{v}\:=\:\mathrm{93} \\ $$$$\mathrm{u}\:=\:\mathrm{1}\:+\:\mathrm{m},\:\:\:\:\mathrm{v}\:=\:\mathrm{92}\:−\:\mathrm{m} \\ $$$$ \\ $$$$\mathrm{from}\:\:\left(\mathrm{ii}\right) \\ $$$$\mathrm{47w}\:−\:\mathrm{211z}\:=\:\mathrm{v}\:\:\:\:\:…..\:\mathrm{ii} \\ $$$$\mathrm{w}\:=\:\mathrm{828}\:−\:\mathrm{9m}\:−\:\mathrm{211n} \\ $$$$\mathrm{z}\:=\:\mathrm{184}\:−\:\mathrm{2m}\:−\:\mathrm{47n} \\ $$$$ \\ $$$$\mathrm{from}\:\:\left(\mathrm{i}\right) \\ $$$$\mathrm{133x}\:+\:\mathrm{65y}\:=\:\mathrm{u}\:\:\:\:\:\:\:……\:\mathrm{i} \\ $$$$\mathrm{x}\:=\:\mathrm{22}\:+\:\mathrm{22m}\:+\:\mathrm{65k} \\ $$$$\mathrm{y}\:=\:−\:\mathrm{45}\:−\:\mathrm{45m}\:−\:\mathrm{133k} \\ $$$$ \\ $$$$\mathrm{Now} \\ $$$$ \\ $$$$\mathrm{x}\:=\:\mathrm{22}\:+\:\mathrm{22m}\:+\:\mathrm{65k} \\ $$$$\mathrm{y}\:=\:−\:\mathrm{45}\:−\:\mathrm{45m}\:−\:\mathrm{133k} \\ $$$$\mathrm{z}\:=\:\mathrm{184}\:−\:\mathrm{2m}\:−\:\mathrm{47n} \\ $$$$\mathrm{w}\:=\:\mathrm{828}\:−\:\mathrm{9m}\:−\:\mathrm{211n} \\ $$$$ \\ $$
Commented by MrW3 last updated on 31/Oct/18
very good!
$${very}\:{good}! \\ $$
Commented by Tawa1 last updated on 31/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MJS last updated on 29/Oct/18
7x+5y+3(5z+4w)=149 5z+4w=v 7x+5y+3v=149 put x=k 5y=149−7k−3v put v=ak+b ⇒ 5y=(149−3b)+(−7−3a)k put a=1 ⇒ 5∣(−7−3a)k put b=−2 ⇒ 5∣(149−3b) v=k−2 ⇒ y=31−2k 5z+4w=k−2 5z=k−2−4w put w=ak+b ⇒ 5z=(−2−4b)+(1−4a)k put a=−1 ⇒ 5∣(1−4a)k put b=2 ⇒ 5∣(−2−4b) w=−k+2 z=k−2 test 7k+5(31−2k)+15(k−2)+12(−k+2)= =7k+155−10k+15k−30−12k+24= =149 general solution x=k y=31−2k z=k−2 w=−k+2
$$\mathrm{7}{x}+\mathrm{5}{y}+\mathrm{3}\left(\mathrm{5}{z}+\mathrm{4}{w}\right)=\mathrm{149} \\ $$$$\mathrm{5}{z}+\mathrm{4}{w}={v} \\ $$$$\mathrm{7}{x}+\mathrm{5}{y}+\mathrm{3}{v}=\mathrm{149} \\ $$$$\mathrm{put}\:{x}={k} \\ $$$$\mathrm{5}{y}=\mathrm{149}−\mathrm{7}{k}−\mathrm{3}{v} \\ $$$$\mathrm{put}\:{v}={ak}+{b}\:\Rightarrow\:\mathrm{5}{y}=\left(\mathrm{149}−\mathrm{3}{b}\right)+\left(−\mathrm{7}−\mathrm{3}{a}\right){k} \\ $$$$\mathrm{put}\:{a}=\mathrm{1}\:\Rightarrow\:\mathrm{5}\mid\left(−\mathrm{7}−\mathrm{3}{a}\right){k} \\ $$$$\mathrm{put}\:{b}=−\mathrm{2}\:\Rightarrow\:\mathrm{5}\mid\left(\mathrm{149}−\mathrm{3}{b}\right) \\ $$$${v}={k}−\mathrm{2} \\ $$$$\Rightarrow\:{y}=\mathrm{31}−\mathrm{2}{k} \\ $$$$\mathrm{5}{z}+\mathrm{4}{w}={k}−\mathrm{2} \\ $$$$\mathrm{5}{z}={k}−\mathrm{2}−\mathrm{4}{w} \\ $$$$\mathrm{put}\:{w}={ak}+{b}\:\Rightarrow\:\mathrm{5}{z}=\left(−\mathrm{2}−\mathrm{4}{b}\right)+\left(\mathrm{1}−\mathrm{4}{a}\right){k} \\ $$$$\mathrm{put}\:{a}=−\mathrm{1}\:\Rightarrow\:\mathrm{5}\mid\left(\mathrm{1}−\mathrm{4}{a}\right){k} \\ $$$$\mathrm{put}\:{b}=\mathrm{2}\:\Rightarrow\:\mathrm{5}\mid\left(−\mathrm{2}−\mathrm{4}{b}\right) \\ $$$${w}=−{k}+\mathrm{2} \\ $$$${z}={k}−\mathrm{2} \\ $$$$\mathrm{test} \\ $$$$\mathrm{7}{k}+\mathrm{5}\left(\mathrm{31}−\mathrm{2}{k}\right)+\mathrm{15}\left({k}−\mathrm{2}\right)+\mathrm{12}\left(−{k}+\mathrm{2}\right)= \\ $$$$=\mathrm{7}{k}+\mathrm{155}−\mathrm{10}{k}+\mathrm{15}{k}−\mathrm{30}−\mathrm{12}{k}+\mathrm{24}= \\ $$$$=\mathrm{149} \\ $$$$ \\ $$$$\mathrm{general}\:\mathrm{solution} \\ $$$${x}={k} \\ $$$${y}=\mathrm{31}−\mathrm{2}{k} \\ $$$${z}={k}−\mathrm{2} \\ $$$${w}=−{k}+\mathrm{2} \\ $$
Commented by MrW3 last updated on 29/Oct/18
thank you sir! but I think what you got represents only a part of all possible solutions. as a general solution it should represent all possible solutions. For example, x=92, y=243, z=−122,w=10 is a solution, but it doesn′t match with the solution scheme you gave. I think for an equation with 4 unknowns we need three parameters to describe the general solution.
$${thank}\:{you}\:{sir}! \\ $$$${but}\:{I}\:{think}\:{what}\:{you}\:{got}\:{represents}\:{only} \\ $$$${a}\:{part}\:{of}\:{all}\:{possible}\:{solutions}.\:{as}\:{a} \\ $$$${general}\:{solution}\:{it}\:{should}\:{represent} \\ $$$${all}\:{possible}\:{solutions}.\: \\ $$$${For}\:{example}, \\ $$$${x}=\mathrm{92},\:{y}=\mathrm{243},\:{z}=−\mathrm{122},{w}=\mathrm{10} \\ $$$${is}\:{a}\:{solution},\:{but}\:{it}\:{doesn}'{t}\:{match}\:{with} \\ $$$${the}\:{solution}\:{scheme}\:{you}\:{gave}. \\ $$$${I}\:{think}\:{for}\:{an}\:{equation}\:{with}\:\mathrm{4}\:{unknowns} \\ $$$${we}\:{need}\:{three}\:{parameters}\:{to}\:{describe} \\ $$$${the}\:{general}\:{solution}. \\ $$
Commented by MJS last updated on 29/Oct/18
you′re right. this was my first attempt to solve an equation like this. now I think we can freely choose 2 variables because the other 2 (if factors are coprime) will do all the work for us in this case choose z and w then it′s always possible to find a matching pair (x, y)
$$\mathrm{you}'\mathrm{re}\:\mathrm{right}.\:\mathrm{this}\:\mathrm{was}\:\mathrm{my}\:\mathrm{first}\:\mathrm{attempt}\:\mathrm{to} \\ $$$$\mathrm{solve}\:\mathrm{an}\:\mathrm{equation}\:\mathrm{like}\:\mathrm{this}.\:\mathrm{now}\:\mathrm{I}\:\mathrm{think}\:\mathrm{we}\:\mathrm{can} \\ $$$$\mathrm{freely}\:\mathrm{choose}\:\mathrm{2}\:\mathrm{variables}\:\mathrm{because}\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{2}\:\left(\mathrm{if}\:\mathrm{factors}\:\mathrm{are}\:\mathrm{coprime}\right)\:\mathrm{will}\:\mathrm{do}\:\mathrm{all}\:\mathrm{the}\:\mathrm{work} \\ $$$$\mathrm{for}\:\mathrm{us} \\ $$$$\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{choose}\:{z}\:\mathrm{and}\:{w}\:\mathrm{then}\:\mathrm{it}'\mathrm{s}\:\mathrm{always} \\ $$$$\mathrm{possible}\:\mathrm{to}\:\mathrm{find}\:\mathrm{a}\:\mathrm{matching}\:\mathrm{pair}\:\left({x},\:{y}\right) \\ $$
Commented by MrW3 last updated on 29/Oct/18
thanks for reviewing sir!
$${thanks}\:{for}\:{reviewing}\:{sir}! \\ $$
Commented by MJS last updated on 29/Oct/18
...in this case 2a+3b+5c+7d=11 we can choose one of these pairs a/b, a/c, a/d, b/c, b/d, c/d in the given case we have 7x+5y+15z+12w we can only choose these x/z, y/z, y/w, z/w does this change the number of parameters? I′m a bit confused...
$$…\mathrm{in}\:\mathrm{this}\:\mathrm{case} \\ $$$$\mathrm{2}{a}+\mathrm{3}{b}+\mathrm{5}{c}+\mathrm{7}{d}=\mathrm{11} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{choose}\:\mathrm{one}\:\mathrm{of}\:\mathrm{these}\:\mathrm{pairs} \\ $$$${a}/{b},\:{a}/{c},\:{a}/{d},\:{b}/{c},\:{b}/{d},\:{c}/{d} \\ $$$$\mathrm{in}\:\mathrm{the}\:\mathrm{given}\:\mathrm{case}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{7}{x}+\mathrm{5}{y}+\mathrm{15}{z}+\mathrm{12}{w} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{choose}\:\mathrm{these} \\ $$$${x}/{z},\:{y}/{z},\:{y}/{w},\:{z}/{w} \\ $$$$\mathrm{does}\:\mathrm{this}\:\mathrm{change}\:\mathrm{the}\:\mathrm{number}\:\mathrm{of}\:\mathrm{parameters}? \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{a}\:\mathrm{bit}\:\mathrm{confused}… \\ $$
Commented by Tawa1 last updated on 29/Oct/18
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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