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Question-133337




Question Number 133337 by mohammad17 last updated on 21/Feb/21
Commented by Dwaipayan Shikari last updated on 21/Feb/21
x=3 y=1 z=2
$${x}=\mathrm{3}\:{y}=\mathrm{1}\:{z}=\mathrm{2} \\ $$
Commented by mohammad17 last updated on 21/Feb/21
can you give me stebs
$${can}\:{you}\:{give}\:{me}\:{stebs} \\ $$
Commented by Dwaipayan Shikari last updated on 21/Feb/21
x+2y+3z=11 →(a)  2x−2y−z=2 →(b)  a+b⇒3x+2z=13  3x−y+2z=12 →(c)⇒y=1  x+3z=9→(d)  x=3,  z=2
$${x}+\mathrm{2}{y}+\mathrm{3}{z}=\mathrm{11}\:\rightarrow\left({a}\right) \\ $$$$\mathrm{2}{x}−\mathrm{2}{y}−{z}=\mathrm{2}\:\rightarrow\left({b}\right) \\ $$$${a}+{b}\Rightarrow\mathrm{3}{x}+\mathrm{2}{z}=\mathrm{13} \\ $$$$\mathrm{3}{x}−{y}+\mathrm{2}{z}=\mathrm{12}\:\rightarrow\left({c}\right)\Rightarrow{y}=\mathrm{1} \\ $$$${x}+\mathrm{3}{z}=\mathrm{9}\rightarrow\left({d}\right) \\ $$$${x}=\mathrm{3},\:\:{z}=\mathrm{2} \\ $$
Commented by mohammad17 last updated on 21/Feb/21
please sir help me
$${please}\:{sir}\:{help}\:{me} \\ $$
Commented by bramlexs22 last updated on 21/Feb/21
this not triangular decomposition
$$\mathrm{this}\:\mathrm{not}\:\mathrm{triangular}\:\mathrm{decomposition} \\ $$
Answered by bramlexs22 last updated on 21/Feb/21
(1) 3x−y+2z = 12   (2)x+2y+3z = 11...×(3)  subtract (1) ∧(2)  ⇒ −7y−7z= −21...(4)              −y−z = −3  (2)×2⇒2x+4y+6z = 22  (3)       ⇒2x−2y−z = 2  subtract (2)∧(3)  ⇒6y+7z = 20 ...(5)  (4)×6⇒−6y−6z = −18   (5)     ⇒     6y +7z = 20                                    z = 2    then  { ((3x−y+2z = 12)),((          y+z   = 3)),((                 z  = 2)) :}   we get  { ((y+2 = 3 →y=1)),((3x−1+4 = 12 ; x= 3)) :}  solution set { 3,1,2}
$$\left(\mathrm{1}\right)\:\mathrm{3x}−\mathrm{y}+\mathrm{2z}\:=\:\mathrm{12}\: \\ $$$$\left(\mathrm{2}\right)\mathrm{x}+\mathrm{2y}+\mathrm{3z}\:=\:\mathrm{11}…×\left(\mathrm{3}\right) \\ $$$$\mathrm{subtract}\:\left(\mathrm{1}\right)\:\wedge\left(\mathrm{2}\right) \\ $$$$\Rightarrow\:−\mathrm{7y}−\mathrm{7z}=\:−\mathrm{21}…\left(\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:−\mathrm{y}−\mathrm{z}\:=\:−\mathrm{3} \\ $$$$\left(\mathrm{2}\right)×\mathrm{2}\Rightarrow\mathrm{2x}+\mathrm{4y}+\mathrm{6z}\:=\:\mathrm{22} \\ $$$$\left(\mathrm{3}\right)\:\:\:\:\:\:\:\Rightarrow\mathrm{2x}−\mathrm{2y}−\mathrm{z}\:=\:\mathrm{2} \\ $$$$\mathrm{subtract}\:\left(\mathrm{2}\right)\wedge\left(\mathrm{3}\right) \\ $$$$\Rightarrow\mathrm{6y}+\mathrm{7z}\:=\:\mathrm{20}\:…\left(\mathrm{5}\right) \\ $$$$\left(\mathrm{4}\right)×\mathrm{6}\Rightarrow−\mathrm{6y}−\mathrm{6z}\:=\:−\mathrm{18} \\ $$$$\:\left(\mathrm{5}\right)\:\:\:\:\:\Rightarrow\:\:\:\:\:\mathrm{6y}\:+\mathrm{7z}\:=\:\mathrm{20} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\:=\:\mathrm{2} \\ $$$$ \\ $$$$\mathrm{then}\:\begin{cases}{\mathrm{3x}−\mathrm{y}+\mathrm{2z}\:=\:\mathrm{12}}\\{\:\:\:\:\:\:\:\:\:\:\mathrm{y}+\mathrm{z}\:\:\:=\:\mathrm{3}}\\{\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{z}\:\:=\:\mathrm{2}}\end{cases} \\ $$$$\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{y}+\mathrm{2}\:=\:\mathrm{3}\:\rightarrow\mathrm{y}=\mathrm{1}}\\{\mathrm{3x}−\mathrm{1}+\mathrm{4}\:=\:\mathrm{12}\:;\:\mathrm{x}=\:\mathrm{3}}\end{cases} \\ $$$$\mathrm{solution}\:\mathrm{set}\:\left\{\:\mathrm{3},\mathrm{1},\mathrm{2}\right\} \\ $$

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