Question-210156

Question Number 210156 by Spillover last updated on 01/Aug/24

Answered by A5T last updated on 01/Aug/24

(i)+2×(ii)⇒x=((8−7z)/7)=(8/7)−z (iii)+(i)⇒7x+(a^2 −9)z=a+4 ⇒x=((−(a^2 −9)z+a+4)/7) ⇒((4−a)/7)=(((16−a^2 )z)/7)⇒z=((4−a)/(16−a^2 ))=(1/(4+a))( OR a=4) (i)⇒4+a=0⇒a=−4 (ii) x=(8/7)−(1/(4+a))=((25+8a)/(7(4+a)))⇒((56+10a)/(28+7a))=y ⇒(x,y,z)=(((25+8a)/(28+7a)),((56+10a)/(28+7a)),(1/(4+a))) ⇒when a≠4, Q\4, then (x,y,z) is unique (iii) a=4⇒4x+y+2z=6...(iii)′ (iii)′−(ii)⇒3x−y+5z=2...(ii) 3×(i)−(ii)⇒7y−14z=10⇒y=2z+((10)/7) ⇒x=((−28z+32)/(28))=−z+(8/7) a=4⇒(x,y,z)=(−z+(8/7),2z+((10)/7),z) Hence, infinitely many solutions when a=4.

$$\left({i}\right)+\mathrm{2}×\left({ii}\right)\Rightarrow{x}=\frac{\mathrm{8}−\mathrm{7}{z}}{\mathrm{7}}=\frac{\mathrm{8}}{\mathrm{7}}−{z} \\ $$$$\left({iii}\right)+\left({i}\right)\Rightarrow\mathrm{7}{x}+\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}={a}+\mathrm{4} \\ $$$$\Rightarrow{x}=\frac{−\left({a}^{\mathrm{2}} −\mathrm{9}\right){z}+{a}+\mathrm{4}}{\mathrm{7}} \\ $$$$\Rightarrow\frac{\mathrm{4}−{a}}{\mathrm{7}}=\frac{\left(\mathrm{16}−{a}^{\mathrm{2}} \right){z}}{\mathrm{7}}\Rightarrow{z}=\frac{\mathrm{4}−{a}}{\mathrm{16}−{a}^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{4}+{a}}\left(\:{OR}\:{a}=\mathrm{4}\right) \\ $$$$\left({i}\right)\Rightarrow\mathrm{4}+{a}=\mathrm{0}\Rightarrow{a}=−\mathrm{4} \\ $$$$\left({ii}\right)\:{x}=\frac{\mathrm{8}}{\mathrm{7}}−\frac{\mathrm{1}}{\mathrm{4}+{a}}=\frac{\mathrm{25}+\mathrm{8}{a}}{\mathrm{7}\left(\mathrm{4}+{a}\right)}\Rightarrow\frac{\mathrm{56}+\mathrm{10}{a}}{\mathrm{28}+\mathrm{7}{a}}={y} \\ $$$$\Rightarrow\left({x},{y},{z}\right)=\left(\frac{\mathrm{25}+\mathrm{8}{a}}{\mathrm{28}+\mathrm{7}{a}},\frac{\mathrm{56}+\mathrm{10}{a}}{\mathrm{28}+\mathrm{7}{a}},\frac{\mathrm{1}}{\mathrm{4}+{a}}\right) \\ $$$$\:\Rightarrow{when}\:{a}\neq\mathrm{4},\:\mathbb{Q}\backslash\mathrm{4},\:{then}\:\left({x},{y},{z}\right)\:{is}\:{unique} \\ $$$$ \\ $$$$\left({iii}\right)\:\:\:\:\:\:{a}=\mathrm{4}\Rightarrow\mathrm{4}{x}+{y}+\mathrm{2}{z}=\mathrm{6}…\left({iii}\right)' \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\left({iii}\right)'−\left({ii}\right)\Rightarrow\mathrm{3}{x}−{y}+\mathrm{5}{z}=\mathrm{2}…\left({ii}\right) \\ $$$$\mathrm{3}×\left({i}\right)−\left({ii}\right)\Rightarrow\mathrm{7}{y}−\mathrm{14}{z}=\mathrm{10}\Rightarrow{y}=\mathrm{2}{z}+\frac{\mathrm{10}}{\mathrm{7}} \\ $$$$\Rightarrow{x}=\frac{−\mathrm{28}{z}+\mathrm{32}}{\mathrm{28}}=−{z}+\frac{\mathrm{8}}{\mathrm{7}} \\ $$$${a}=\mathrm{4}\Rightarrow\left({x},{y},{z}\right)=\left(−{z}+\frac{\mathrm{8}}{\mathrm{7}},\mathrm{2}{z}+\frac{\mathrm{10}}{\mathrm{7}},{z}\right) \\ $$$${Hence},\:{infinitely}\:{many}\:{solutions}\:{when}\:{a}=\mathrm{4}. \\ $$

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