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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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