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Question Number 87050 by jagoll last updated on 02/Apr/20
what are the roots  of the   system of equation (x/y)+(y/(x+1)) = (4/3)  and x+y + xy = 5 ?
$$\mathrm{what}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\:\mathrm{of}\:\mathrm{the}\: \\ $$$$\mathrm{system}\:\mathrm{of}\:\mathrm{equation}\:\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}+\mathrm{1}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{and}\:\mathrm{x}+\mathrm{y}\:+\:\mathrm{xy}\:=\:\mathrm{5}\:? \\ $$
Commented by john santu last updated on 02/Apr/20
⇒x+y+xy+1 = 6 ⇒x(y+1)=5−y  x = ((5−y)/(y+1))  (x+1)(y+1) = 6  x+1 = (6/(y+1)) ⇒ ((5−y)/(y(y+1)))+((y(y+1))/6) = (4/3)  30−6y+y^2 (y+1)^2 = 8y(y+1)  y^4 +2y^3 +y^2 −8y^2 −14y+30=0  y^4 +2y^3 −7y^2 −14y+30=0
$$\Rightarrow{x}+{y}+{xy}+\mathrm{1}\:=\:\mathrm{6}\:\Rightarrow{x}\left({y}+\mathrm{1}\right)=\mathrm{5}−{y} \\ $$$${x}\:=\:\frac{\mathrm{5}−{y}}{{y}+\mathrm{1}} \\ $$$$\left({x}+\mathrm{1}\right)\left({y}+\mathrm{1}\right)\:=\:\mathrm{6} \\ $$$${x}+\mathrm{1}\:=\:\frac{\mathrm{6}}{{y}+\mathrm{1}}\:\Rightarrow\:\frac{\mathrm{5}−{y}}{{y}\left({y}+\mathrm{1}\right)}+\frac{{y}\left({y}+\mathrm{1}\right)}{\mathrm{6}}\:=\:\frac{\mathrm{4}}{\mathrm{3}} \\ $$$$\mathrm{30}−\mathrm{6}{y}+{y}^{\mathrm{2}} \left({y}+\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{8}{y}\left({y}+\mathrm{1}\right) \\ $$$${y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} +{y}^{\mathrm{2}} −\mathrm{8}{y}^{\mathrm{2}} −\mathrm{14}{y}+\mathrm{30}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} +\mathrm{2}{y}^{\mathrm{3}} −\mathrm{7}{y}^{\mathrm{2}} −\mathrm{14}{y}+\mathrm{30}=\mathrm{0} \\ $$$$ \\ $$
Commented by MJS last updated on 02/Apr/20
no “nice” solution  y≈1.75925±.436230i∨y≈−2.75925±1.23214i
$$\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$$${y}\approx\mathrm{1}.\mathrm{75925}\pm.\mathrm{436230i}\vee{y}\approx−\mathrm{2}.\mathrm{75925}\pm\mathrm{1}.\mathrm{23214i} \\ $$
Commented by jagoll last updated on 02/Apr/20
thank you mister
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{mister} \\ $$

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