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Question Number 59442 by Tawa1 last updated on 10/May/19

Solve the system xy + 3x + 2y = − 6 ..... (i) yx + y + 3z = − 3 ..... (ii) zx + 2z + x = 2 ..... (iii)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{system} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{xy}\:+\:\mathrm{3x}\:+\:\mathrm{2y}\:\:=\:−\:\mathrm{6}\:\:\:\:\:\:\:.....\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{yx}\:+\:\mathrm{y}\:+\:\mathrm{3z}\:\:\:=\:−\:\mathrm{3}\:\:\:\:\:\:\:\:\:\:.....\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{zx}\:+\:\mathrm{2z}\:+\:\mathrm{x}\:\:\:=\:\:\mathrm{2}\:\:\:\:\:\:\:\:\:\:.....\:\left(\mathrm{iii}\right) \\ $$

Commented by mr W last updated on 10/May/19

yz+y+3z=−3 ...(ii) ????

$${yz}+{y}+\mathrm{3}{z}=−\mathrm{3}\:\:...\left({ii}\right)\:\:\:\:???? \\ $$

Commented by Tawa1 last updated on 10/May/19

No sir. The question is like that. But you can help me solve it the way you changed it. maybe questiin error. but that is the question

$$\mathrm{No}\:\mathrm{sir}.\:\:\mathrm{The}\:\mathrm{question}\:\mathrm{is}\:\mathrm{like}\:\mathrm{that}. \\ $$$$\:\:\:\:\mathrm{But}\:\mathrm{you}\:\mathrm{can}\:\mathrm{help}\:\mathrm{me}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{the}\:\mathrm{way}\:\mathrm{you}\:\mathrm{changed}\:\mathrm{it}. \\ $$$$\mathrm{maybe}\:\mathrm{questiin}\:\mathrm{error}.\:\mathrm{but}\:\mathrm{that}\:\mathrm{is}\:\mathrm{the}\:\mathrm{question} \\ $$$$ \\ $$

Answered by ajfour last updated on 10/May/19

z=((2−x)/(2+x)) yx+y+3(((2−x)/(2+x)))=−3 ⇒ yx+y=((12)/(2+x)) from (i) y=−3(((x+2)/(x+2))) so assuming x≠−2 y=−3 ⇒ ((12)/(2+x))= −3x−3 or 4+(x+1)(2+x)=0 ⇒ x^2 +3x+6 =0 (x+(3/2))^2 =−6+(9/4) x=−(3/2)± ((i(√(15)))/2) z=((2−x)/(2+x)) .

$$\:\:\mathrm{z}=\frac{\mathrm{2}−\mathrm{x}}{\mathrm{2}+\mathrm{x}} \\ $$$$\:\:\:\:\mathrm{yx}+\mathrm{y}+\mathrm{3}\left(\frac{\mathrm{2}−\mathrm{x}}{\mathrm{2}+\mathrm{x}}\right)=−\mathrm{3} \\ $$$$\Rightarrow\:\:\:\mathrm{yx}+\mathrm{y}=\frac{\mathrm{12}}{\mathrm{2}+\mathrm{x}} \\ $$$$\mathrm{from}\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{y}=−\mathrm{3}\left(\frac{\mathrm{x}+\mathrm{2}}{\mathrm{x}+\mathrm{2}}\right)\:\:\:\:\: \\ $$$$\mathrm{so}\:\mathrm{assuming}\:\:\mathrm{x}\neq−\mathrm{2} \\ $$$$\:\:\:\:\:\mathrm{y}=−\mathrm{3} \\ $$$$\Rightarrow\:\:\:\:\frac{\mathrm{12}}{\mathrm{2}+\mathrm{x}}=\:−\mathrm{3x}−\mathrm{3} \\ $$$$\mathrm{or}\:\:\:\:\:\:\:\mathrm{4}+\left(\mathrm{x}+\mathrm{1}\right)\left(\mathrm{2}+\mathrm{x}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{3x}+\mathrm{6}\:=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\:\:\left(\mathrm{x}+\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} =−\mathrm{6}+\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\mathrm{x}=−\frac{\mathrm{3}}{\mathrm{2}}\pm\:\frac{\mathrm{i}\sqrt{\mathrm{15}}}{\mathrm{2}} \\ $$$$\:\:\:\:\mathrm{z}=\frac{\mathrm{2}−\mathrm{x}}{\mathrm{2}+\mathrm{x}}\:. \\ $$

Commented by Tawa1 last updated on 17/May/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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