Question-116192

Question Number 116192 by bemath last updated on 01/Oct/20

Commented by MJS_new last updated on 01/Oct/20

x=y=z=0 not sure how to further solve it

$${x}={y}={z}=\mathrm{0} \\ $$$$\mathrm{not}\:\mathrm{sure}\:\mathrm{how}\:\mathrm{to}\:\mathrm{further}\:\mathrm{solve}\:\mathrm{it} \\ $$

Answered by TANMAY PANACEA last updated on 01/Oct/20

(1+xy)((1/x)+(1/y))=5 (1/x)+(1/y)+x+y=5 (1/y)+(1/z)+y+z=6 (1/z)+(1/x)+z+x=7 x+y+z+(1/x)+(1/y)+(1/z)=((18)/2)=9 z+(1/z)=4→z^2 −4z+1=0 z=((4±(√(16−4)))/2)=((4±2(√3))/2)=2±(√3) y+(1/y)=2→(y−1)^2 =0→y=1,1 x+(1/x)=3→x^2 −3x+1=0 x=((3±(√(9−4)))/2)=((3±(√5))/2)

$$\left(\mathrm{1}+{xy}\right)\left(\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}\right)=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+{x}+{y}=\mathrm{5} \\ $$$$\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}+{y}+{z}=\mathrm{6} \\ $$$$\frac{\mathrm{1}}{{z}}+\frac{\mathrm{1}}{{x}}+{z}+{x}=\mathrm{7} \\ $$$${x}+{y}+{z}+\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}+\frac{\mathrm{1}}{{z}}=\frac{\mathrm{18}}{\mathrm{2}}=\mathrm{9} \\ $$$${z}+\frac{\mathrm{1}}{{z}}=\mathrm{4}\rightarrow{z}^{\mathrm{2}} −\mathrm{4}{z}+\mathrm{1}=\mathrm{0} \\ $$$${z}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{4}\pm\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{2}}=\mathrm{2}\pm\sqrt{\mathrm{3}}\: \\ $$$${y}+\frac{\mathrm{1}}{{y}}=\mathrm{2}\rightarrow\left({y}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{0}\rightarrow{y}=\mathrm{1},\mathrm{1} \\ $$$${x}+\frac{\mathrm{1}}{{x}}=\mathrm{3}\rightarrow{x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{1}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{3}\pm\sqrt{\mathrm{9}−\mathrm{4}}}{\mathrm{2}}=\frac{\mathrm{3}\pm\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$

Commented by MJS_new last updated on 01/Oct/20