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

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

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

$$yz+y+3z=-3...\left(ii\right)????$$

Commented by Tawa1 last updated on 10/May/19

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

Answered by ajfour last updated on 10/May/19

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

Commented by Tawa1 last updated on 17/May/19

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

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