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Question Number 8933 by Hariom maurya last updated on 06/Nov/16

3x+3y+2z=1,x+2y=4,10y+3z=−2, 2x−3y−z=5

$$\mathrm{3x}+\mathrm{3y}+\mathrm{2z}=\mathrm{1},\mathrm{x}+\mathrm{2y}=\mathrm{4},\mathrm{10y}+\mathrm{3z}=−\mathrm{2}, \\ $$$$\mathrm{2x}−\mathrm{3y}−\mathrm{z}=\mathrm{5} \\ $$

Answered by Rasheed Soomro last updated on 06/Nov/16

3x+3y+2z=1.............(i) x+2y=4......................(ii) 10y+3z=−2................(iii) 2x−3y−z=5.................(iv) (vi)⇒z=2x−3y−5...........(vii) (i)⇒3x+3y+2(2x−3y−5)=1 ⇒7x−3y=11.............(v) 7×(ii)⇒7x+14y=28.......(vi) (vi)−(v)⇒17y=17⇒y=1 (ii)⇒x=4−2y=4−2(1)=2⇒x=2 (vii)⇒z=2x−3y−5=2(2)−3(1)−5=−4⇒z=−4 x=2, y=1, z=−4 (iii) also satisfied. Note: Three equations are sufficient for solution.So in this solution (iii) has not been used.

$$\mathrm{3x}+\mathrm{3y}+\mathrm{2z}=\mathrm{1}.............\left(\mathrm{i}\right) \\ $$$$\mathrm{x}+\mathrm{2y}=\mathrm{4}......................\left(\mathrm{ii}\right) \\ $$$$\mathrm{10y}+\mathrm{3z}=−\mathrm{2}................\left(\mathrm{iii}\right) \\ $$$$\mathrm{2x}−\mathrm{3y}−\mathrm{z}=\mathrm{5}.................\left(\mathrm{iv}\right) \\ $$$$\left(\mathrm{vi}\right)\Rightarrow\mathrm{z}=\mathrm{2x}−\mathrm{3y}−\mathrm{5}...........\left(\mathrm{vii}\right) \\ $$$$\left(\mathrm{i}\right)\Rightarrow\mathrm{3x}+\mathrm{3y}+\mathrm{2}\left(\mathrm{2x}−\mathrm{3y}−\mathrm{5}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\Rightarrow\mathrm{7x}−\mathrm{3y}=\mathrm{11}.............\left(\mathrm{v}\right) \\ $$$$\mathrm{7}×\left(\mathrm{ii}\right)\Rightarrow\mathrm{7x}+\mathrm{14y}=\mathrm{28}.......\left(\mathrm{vi}\right) \\ $$$$\left(\mathrm{vi}\right)−\left(\mathrm{v}\right)\Rightarrow\mathrm{17y}=\mathrm{17}\Rightarrow\mathrm{y}=\mathrm{1} \\ $$$$\left(\mathrm{ii}\right)\Rightarrow\mathrm{x}=\mathrm{4}−\mathrm{2y}=\mathrm{4}−\mathrm{2}\left(\mathrm{1}\right)=\mathrm{2}\Rightarrow\mathrm{x}=\mathrm{2} \\ $$$$\left(\mathrm{vii}\right)\Rightarrow\mathrm{z}=\mathrm{2x}−\mathrm{3y}−\mathrm{5}=\mathrm{2}\left(\mathrm{2}\right)−\mathrm{3}\left(\mathrm{1}\right)−\mathrm{5}=−\mathrm{4}\Rightarrow\mathrm{z}=−\mathrm{4} \\ $$$$\mathrm{x}=\mathrm{2},\:\mathrm{y}=\mathrm{1},\:\mathrm{z}=−\mathrm{4} \\ $$$$\left(\mathrm{iii}\right)\:\mathrm{also}\:\mathrm{satisfied}. \\ $$$$\mathrm{Note}:\:\mathrm{Three}\:\mathrm{equations}\:\mathrm{are}\:\boldsymbol{\mathrm{sufficient}}\:\mathrm{for} \\ $$$$\mathrm{solution}.\mathrm{So}\:\mathrm{in}\:\mathrm{this}\:\mathrm{solution}\:\left(\mathrm{iii}\right)\:\mathrm{has}\:\boldsymbol{\mathrm{not}}\:\mathrm{been}\:\:\mathrm{used}. \\ $$

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