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Question Number 161946 by mathlove last updated on 24/Dec/21

Answered by Rasheed.Sindhi last updated on 29/Dec/21

x+y(x+1)=3⇒y=((3−x)/(x+1)) ⇒y+1=((3−x+x+1)/(x+1))=(4/(x+1)) y+z(y+1)=5⇒((3−x)/(x+1))+z((4/(x+1)))=5 ⇒3−x+4z=5x+5⇒x=((2z−1)/3) z+x(z+1)=7⇒z+(((2z−1)/3))(z+1)=7 ⇒3z+2z^2 +z−1=21 z^2 +2z−11=0 z=((−2±(√(4+44)))/2)=((−2±4(√3))/2)=−1±2(√3) x=((2z−1)/3)=((2(−1±2(√3) )−1)/3)=((−3±4(√3))/3) y=((3−x)/(x+1))=((3−(((−3±4(√3))/3)))/((((−3±4(√3))/3))+1)) =((9+3∓4(√3))/(±4(√3)))×((±4(√3))/(±4(√3)))=((±48(√3)−48)/(48)) y=−1±(√3) x+y+z=(((−3±4(√3))/3))+(−1±(√3))+(−1±2(√(3 )) ) =((−3±4(√3) −3±3(√3)−3±6(√3) )/3) =((−9±13(√3) )/3) (x+y+z)^2 =(((−9±13(√3) )/3))^2 =((81∓234(√3) +507)/9) =((588∓234(√3) )/9)=((196∓78(√3))/3)

$${x}+{y}\left({x}+\mathrm{1}\right)=\mathrm{3}\Rightarrow{y}=\frac{\mathrm{3}−{x}}{{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Rightarrow{y}+\mathrm{1}=\frac{\mathrm{3}−{x}+{x}+\mathrm{1}}{{x}+\mathrm{1}}=\frac{\mathrm{4}}{{x}+\mathrm{1}} \\ $$$${y}+{z}\left({y}+\mathrm{1}\right)=\mathrm{5}\Rightarrow\frac{\mathrm{3}−{x}}{{x}+\mathrm{1}}+{z}\left(\frac{\mathrm{4}}{{x}+\mathrm{1}}\right)=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\mathrm{3}−{x}+\mathrm{4}{z}=\mathrm{5}{x}+\mathrm{5}\Rightarrow{x}=\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{3}} \\ $$$${z}+{x}\left({z}+\mathrm{1}\right)=\mathrm{7}\Rightarrow{z}+\left(\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{3}}\right)\left({z}+\mathrm{1}\right)=\mathrm{7} \\ $$$$\Rightarrow\mathrm{3}{z}+\mathrm{2}{z}^{\mathrm{2}} +{z}−\mathrm{1}=\mathrm{21} \\ $$$$\:\:\:\:\:\:{z}^{\mathrm{2}} +\mathrm{2}{z}−\mathrm{11}=\mathrm{0} \\ $$$${z}=\frac{−\mathrm{2}\pm\sqrt{\mathrm{4}+\mathrm{44}}}{\mathrm{2}}=\frac{−\mathrm{2}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{2}}=−\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${x}=\frac{\mathrm{2}{z}−\mathrm{1}}{\mathrm{3}}=\frac{\mathrm{2}\left(−\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{3}}\:\right)−\mathrm{1}}{\mathrm{3}}=\frac{−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${y}=\frac{\mathrm{3}−{x}}{{x}+\mathrm{1}}=\frac{\mathrm{3}−\left(\frac{−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)}{\left(\frac{−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)+\mathrm{1}} \\ $$$$\:\:\:\:\:=\frac{\mathrm{9}+\mathrm{3}\mp\mathrm{4}\sqrt{\mathrm{3}}}{\pm\mathrm{4}\sqrt{\mathrm{3}}}×\frac{\pm\mathrm{4}\sqrt{\mathrm{3}}}{\pm\mathrm{4}\sqrt{\mathrm{3}}}=\frac{\pm\mathrm{48}\sqrt{\mathrm{3}}−\mathrm{48}}{\mathrm{48}} \\ $$$$\:\:\:\:{y}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$${x}+{y}+{z}=\left(\frac{−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\right)+\left(−\mathrm{1}\pm\sqrt{\mathrm{3}}\right)+\left(−\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{3}\:}\:\right) \\ $$$$=\frac{−\mathrm{3}\pm\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{3}\pm\mathrm{3}\sqrt{\mathrm{3}}−\mathrm{3}\pm\mathrm{6}\sqrt{\mathrm{3}}\:\:}{\mathrm{3}} \\ $$$$=\frac{−\mathrm{9}\pm\mathrm{13}\sqrt{\mathrm{3}}\:}{\mathrm{3}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left(\frac{−\mathrm{9}\pm\mathrm{13}\sqrt{\mathrm{3}}\:}{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{81}\mp\mathrm{234}\sqrt{\mathrm{3}}\:+\mathrm{507}}{\mathrm{9}} \\ $$$$\:\:\:\:\:\:\:\:=\frac{\mathrm{588}\mp\mathrm{234}\sqrt{\mathrm{3}}\:}{\mathrm{9}}=\frac{\mathrm{196}\mp\mathrm{78}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

T^(H^A N) X_(A_(LO_(T) ^(L) T) ) S I_(I_• ) ^(•_ ) R

$$\:\underset{\underset{\mathrm{L}\underset{\mathrm{T}} {\overset{\mathrm{L}} {\mathrm{O}}T}} {\mathrm{A}}} {\mathbb{T}^{\mathbb{H}^{\mathbb{A}} \mathbb{N}} \mathbb{X}}\: \\ $$$$\:\:\:\mathcal{S}\:\:\:\:\underset{\underset{\bullet} {\mathrm{I}}} {\overset{\underset{} {\bullet}} {\mathrm{I}}}\:\:\:\:\mathcal{R} \\ $$

Answered by mr W last updated on 29/Dec/21

let u=x+1 v=y+1 w=z+1 eqn. (i) becomes u−1+v−1+(u−1)(v−1)=3 uv=4 ...(I) similarly vw=6 ...(II) wu=8 ...(III) (I)×(II)×(III): (uvw)^2 =4×6×8=192 uvw=±(√(192))=±8(√3) ...(IV) (IV)/(I): w=±((8(√3))/4)=±2(√3)=z+1 ⇒z=−1±2(√3) similarly u=±((8(√3))/6)=±((4(√3))/3)=x+1 ⇒x=−1±((4(√3))/3) v=±((8(√3))/8)=±(√3)=y+1 ⇒y=−1±(√3) (x+y+z)^2 =(u+v+w−3)^2 =(±((4(√3))/3)±(√3)±2(√3)−3)^2 =(±((13(√3))/3)−3)^2 = { (((196−78(√3))/3)),(((196+78(√3))/3)) :}

$${let} \\ $$$${u}={x}+\mathrm{1} \\ $$$${v}={y}+\mathrm{1} \\ $$$${w}={z}+\mathrm{1} \\ $$$${eqn}.\:\left({i}\right)\:{becomes} \\ $$$${u}−\mathrm{1}+{v}−\mathrm{1}+\left({u}−\mathrm{1}\right)\left({v}−\mathrm{1}\right)=\mathrm{3} \\ $$$${uv}=\mathrm{4}\:\:\:...\left({I}\right) \\ $$$${similarly} \\ $$$${vw}=\mathrm{6}\:\:\:...\left({II}\right) \\ $$$${wu}=\mathrm{8}\:\:\:...\left({III}\right) \\ $$$$\left({I}\right)×\left({II}\right)×\left({III}\right): \\ $$$$\left({uvw}\right)^{\mathrm{2}} =\mathrm{4}×\mathrm{6}×\mathrm{8}=\mathrm{192} \\ $$$${uvw}=\pm\sqrt{\mathrm{192}}=\pm\mathrm{8}\sqrt{\mathrm{3}}\:\:\:...\left({IV}\right) \\ $$$$\left({IV}\right)/\left({I}\right): \\ $$$${w}=\pm\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{4}}=\pm\mathrm{2}\sqrt{\mathrm{3}}={z}+\mathrm{1}\:\Rightarrow{z}=−\mathrm{1}\pm\mathrm{2}\sqrt{\mathrm{3}} \\ $$$${similarly} \\ $$$${u}=\pm\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{6}}=\pm\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}={x}+\mathrm{1}\:\Rightarrow{x}=−\mathrm{1}\pm\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$$${v}=\pm\frac{\mathrm{8}\sqrt{\mathrm{3}}}{\mathrm{8}}=\pm\sqrt{\mathrm{3}}={y}+\mathrm{1}\:\Rightarrow{y}=−\mathrm{1}\pm\sqrt{\mathrm{3}} \\ $$$$\left({x}+{y}+{z}\right)^{\mathrm{2}} =\left({u}+{v}+{w}−\mathrm{3}\right)^{\mathrm{2}} =\left(\pm\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\mathrm{3}}\pm\sqrt{\mathrm{3}}\pm\mathrm{2}\sqrt{\mathrm{3}}−\mathrm{3}\right)^{\mathrm{2}} \\ $$$$=\left(\pm\frac{\mathrm{13}\sqrt{\mathrm{3}}}{\mathrm{3}}−\mathrm{3}\right)^{\mathrm{2}} =\begin{cases}{\frac{\mathrm{196}−\mathrm{78}\sqrt{\mathrm{3}}}{\mathrm{3}}}\\{\frac{\mathrm{196}+\mathrm{78}\sqrt{\mathrm{3}}}{\mathrm{3}}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

Sir I didn′t understand last line.

$$\mathrm{Sir}\:\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{understand}\:\mathrm{last}\:\mathrm{line}. \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

SMART METHOD SiR!

$$\mathbb{SMART}\:\:\mathcal{METHOD}\:\:\mathrm{SiR}! \\ $$

Commented by mr W last updated on 29/Dec/21

it only shows that there are two values.

$${it}\:{only}\:{shows}\:{that}\:{there}\:{are}\:{two} \\ $$$${values}. \\ $$

Commented by Rasheed.Sindhi last updated on 29/Dec/21

Again Thanx!

$$\mathrm{Again}\:\mathrm{Thanx}! \\ $$

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