x-10y-50z-500-x-y-z-100-find-x-y-z-

Question Number 86741 by M±th+et£s last updated on 30/Mar/20

$$\begin{cases}{{x}+\mathrm{10}{y}+\mathrm{50}{z}=\mathrm{500}}\\{{x}+{y}+{z}=\mathrm{100}}\end{cases} \\ $$$$ \\ $$$${find}\:{x},{y},{z} \\ $$

Commented by mr W last updated on 30/Mar/20

$${x},{y},{z}\in{N}\:? \\ $$

Commented by M±th+et£s last updated on 30/Mar/20

$${yes}\:{sir} \\ $$

Commented by mr W last updated on 30/Mar/20

$${yes},\:{there}\:{is}\:{this}\:{one}\:{solution}\:{for}\:{N}. \\ $$

Commented by M±th+et£s last updated on 30/Mar/20

$${x}=\mathrm{60} \\ $$$${y}=\mathrm{39} \\ $$$${z}=\mathrm{1} \\ $$

Answered by mr W last updated on 30/Mar/20

x+10y+50z=500 x=10x′ 10x′+10y+50z=500 ⇒x′+y+5z=50 x′+y=1 ⇒x′=−k ⇒y=k+1 u+5z=50 ⇒u=−5h ⇒z=h+10 x′=(−5h)(−k)=5hk=n ⇒x=10n y=(−5h)(k+1)=−n−5h 10n−n−5h+h+10=100 9n−4h=90 ⇒n=4k+6 ⇒h=9k−9 general solution: ⇒x=20(2k+3) ⇒y=39−49k ⇒z=9k+1 for x,y,z∈N k=0: x=60, y=39, z=1

$${x}+\mathrm{10}{y}+\mathrm{50}{z}=\mathrm{500} \\ $$$${x}=\mathrm{10}{x}' \\ $$$$\mathrm{10}{x}'+\mathrm{10}{y}+\mathrm{50}{z}=\mathrm{500} \\ $$$$\Rightarrow{x}'+{y}+\mathrm{5}{z}=\mathrm{50} \\ $$$${x}'+{y}=\mathrm{1} \\ $$$$\Rightarrow{x}'=−{k} \\ $$$$\Rightarrow{y}={k}+\mathrm{1} \\ $$$${u}+\mathrm{5}{z}=\mathrm{50} \\ $$$$\Rightarrow{u}=−\mathrm{5}{h} \\ $$$$\Rightarrow{z}={h}+\mathrm{10} \\ $$$${x}'=\left(−\mathrm{5}{h}\right)\left(−{k}\right)=\mathrm{5}{hk}={n}\:\Rightarrow{x}=\mathrm{10}{n} \\ $$$${y}=\left(−\mathrm{5}{h}\right)\left({k}+\mathrm{1}\right)=−{n}−\mathrm{5}{h} \\ $$$$ \\ $$$$\mathrm{10}{n}−{n}−\mathrm{5}{h}+{h}+\mathrm{10}=\mathrm{100} \\ $$$$\mathrm{9}{n}−\mathrm{4}{h}=\mathrm{90} \\ $$$$\Rightarrow{n}=\mathrm{4}{k}+\mathrm{6} \\ $$$$\Rightarrow{h}=\mathrm{9}{k}−\mathrm{9} \\ $$$$ \\ $$$${general}\:{solution}: \\ $$$$\Rightarrow{x}=\mathrm{20}\left(\mathrm{2}{k}+\mathrm{3}\right) \\ $$$$\Rightarrow{y}=\mathrm{39}−\mathrm{49}{k} \\ $$$$\Rightarrow{z}=\mathrm{9}{k}+\mathrm{1} \\ $$$${for}\:{x},{y},{z}\in{N} \\ $$$${k}=\mathrm{0}: \\ $$$${x}=\mathrm{60},\:{y}=\mathrm{39},\:{z}=\mathrm{1} \\ $$

Commented by M±th+et£s last updated on 30/Mar/20