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x-y-z-1-42x-44y-30z-42-x-y-z-1-0-0-yes-but-solution-




Question Number 209436 by hardmath last updated on 10/Jul/24
 { ((x + y + z = 1)),((42x + 44y + 30z = 42)) :}  (x,y,z)=(1,0,0) yes, but solution...
$$\begin{cases}{\mathrm{x}\:+\:\mathrm{y}\:+\:\mathrm{z}\:=\:\mathrm{1}}\\{\mathrm{42x}\:+\:\mathrm{44y}\:+\:\mathrm{30z}\:=\:\mathrm{42}}\end{cases} \\ $$$$\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\left(\mathrm{1},\mathrm{0},\mathrm{0}\right)\:\mathrm{yes},\:\mathrm{but}\:\mathrm{solution}… \\ $$
Commented by mr W last updated on 10/Jul/24
you mean integer solutions.
$${you}\:{mean}\:{integer}\:{solutions}. \\ $$
Commented by hardmath last updated on 10/Jul/24
yes dear professor
$$\mathrm{yes}\:\mathrm{dear}\:\mathrm{professor} \\ $$
Answered by mr W last updated on 10/Jul/24
z=1−x−y  42x+44y+30z=42  21x+22y+15z=21  21x+22y+15(1−x−y)=21  ⇒7y=6(1−x)  ⇒1−x=7k ⇒x=1−7k  ⇒y=6k  ⇒z=1−1+7k−6k=k  general solution with k∈Z:  x=1−7k  y=6k  z=k
$${z}=\mathrm{1}−{x}−{y} \\ $$$$\mathrm{42}{x}+\mathrm{44}{y}+\mathrm{30}{z}=\mathrm{42} \\ $$$$\mathrm{21}{x}+\mathrm{22}{y}+\mathrm{15}{z}=\mathrm{21} \\ $$$$\mathrm{21}{x}+\mathrm{22}{y}+\mathrm{15}\left(\mathrm{1}−{x}−{y}\right)=\mathrm{21} \\ $$$$\Rightarrow\mathrm{7}{y}=\mathrm{6}\left(\mathrm{1}−{x}\right) \\ $$$$\Rightarrow\mathrm{1}−{x}=\mathrm{7}{k}\:\Rightarrow{x}=\mathrm{1}−\mathrm{7}{k} \\ $$$$\Rightarrow{y}=\mathrm{6}{k} \\ $$$$\Rightarrow{z}=\mathrm{1}−\mathrm{1}+\mathrm{7}{k}−\mathrm{6}{k}={k} \\ $$$${general}\:{solution}\:{with}\:{k}\in{Z}: \\ $$$${x}=\mathrm{1}−\mathrm{7}{k} \\ $$$${y}=\mathrm{6}{k} \\ $$$${z}={k} \\ $$
Commented by hardmath last updated on 13/Jul/24
thankyou dearprofessor
$$\boldsymbol{\mathrm{t}}\mathrm{hank}\boldsymbol{\mathrm{y}}\mathrm{ou}\:\boldsymbol{\mathrm{d}}\mathrm{ear}\boldsymbol{\mathrm{p}}\mathrm{rofessor} \\ $$

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