Menu Close

Question-65087

Tinku Tara 0 Comments
Pin




Question Number 65087 by Tawa1 last updated on 25/Jul/19
Answered by MJS last updated on 25/Jul/19
easier than it seems (1) x+y+z=0 ⇒ z=−x−y insert this in (2) and (3) (2) −3xy(x+y)=18 (3) −7xy(x+y)(x^2 +xy+y^2 )^2 =2058 now simply divide (2) by (−3) (2) xy(x+y)=−6 insert this in (3) (3) 42(x^2 +xy+y^2 )=2058 (x^2 +xy+y^2 )=49 x^2 +xy+y^2 =7 xy+y^2 =7−x^2 x+y=((7−x^2 )/y) now remember xy(x+y)=−6 ⇒ x+y=−(6/(xy)) −(6/(xy))=((7−x^2 )/y) x^3 −7x−6=0 ⇒ x=−2∨x=−1∨x=3 because in all given equations x, y, z are interchangeable we get ((x),(y),(z) ) ∈{ (((−2)),((−1)),(3) ) , (((−2)),(3),((−1)) ) , (((−1)),((−2)),(3) ) , (((−1)),(3),((−2)) ) , ((3),((−2)),((−1)) ) , ((3),((−1)),((−2)) ) }
$$\mathrm{easier}\:\mathrm{than}\:\mathrm{it}\:\mathrm{seems} \\ $$$$\left(\mathrm{1}\right)\:\:{x}+{y}+{z}=\mathrm{0}\:\Rightarrow\:{z}=−{x}−{y} \\ $$$$\mathrm{insert}\:\mathrm{this}\:\mathrm{in}\:\left(\mathrm{2}\right)\:\mathrm{and}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\:−\mathrm{3}{xy}\left({x}+{y}\right)=\mathrm{18} \\ $$$$\left(\mathrm{3}\right)\:\:−\mathrm{7}{xy}\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)^{\mathrm{2}} =\mathrm{2058} \\ $$$$\mathrm{now}\:\mathrm{simply}\:\mathrm{divide}\:\left(\mathrm{2}\right)\:\mathrm{by}\:\left(−\mathrm{3}\right) \\ $$$$\left(\mathrm{2}\right)\:\:{xy}\left({x}+{y}\right)=−\mathrm{6} \\ $$$$\mathrm{insert}\:\mathrm{this}\:\mathrm{in}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{3}\right)\:\:\mathrm{42}\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)=\mathrm{2058} \\ $$$$\:\:\:\:\:\:\:\:\left({x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} \right)=\mathrm{49} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} +{xy}+{y}^{\mathrm{2}} =\mathrm{7} \\ $$$$\:\:\:\:\:\:\:\:{xy}+{y}^{\mathrm{2}} =\mathrm{7}−{x}^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:{x}+{y}=\frac{\mathrm{7}−{x}^{\mathrm{2}} }{{y}} \\ $$$$\mathrm{now}\:\mathrm{remember}\:{xy}\left({x}+{y}\right)=−\mathrm{6}\:\Rightarrow\:{x}+{y}=−\frac{\mathrm{6}}{{xy}} \\ $$$$\:\:\:\:\:\:\:\:−\frac{\mathrm{6}}{{xy}}=\frac{\mathrm{7}−{x}^{\mathrm{2}} }{{y}} \\ $$$$\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} −\mathrm{7}{x}−\mathrm{6}=\mathrm{0} \\ $$$$\:\:\:\:\:\:\:\:\Rightarrow\:{x}=−\mathrm{2}\vee{x}=−\mathrm{1}\vee{x}=\mathrm{3} \\ $$$$\mathrm{because}\:\mathrm{in}\:\mathrm{all}\:\mathrm{given}\:\mathrm{equations}\:{x},\:{y},\:{z}\:\mathrm{are} \\ $$$$\mathrm{interchangeable}\:\mathrm{we}\:\mathrm{get} \\ $$$$\begin{pmatrix}{{x}}\\{{y}}\\{{z}}\end{pmatrix}\:\in\left\{\begin{pmatrix}{−\mathrm{2}}\\{−\mathrm{1}}\\{\mathrm{3}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{2}}\\{\mathrm{3}}\\{−\mathrm{1}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{1}}\\{−\mathrm{2}}\\{\mathrm{3}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{1}}\\{\mathrm{3}}\\{−\mathrm{2}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{3}}\\{−\mathrm{2}}\\{−\mathrm{1}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{3}}\\{−\mathrm{1}}\\{−\mathrm{2}}\end{pmatrix}\:\right\} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
God bless you sir
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
Sir i don′t understand from line 2 and 3 only.
$$\mathrm{Sir}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{from}\:\mathrm{line}\:\mathrm{2}\:\mathrm{and}\:\mathrm{3}\:\mathrm{only}. \\ $$
Commented by MJS last updated on 25/Jul/19
I inserted some explanations
$$\mathrm{I}\:\mathrm{inserted}\:\mathrm{some}\:\mathrm{explanations} \\ $$
Commented by Tawa1 last updated on 25/Jul/19
Ohh, God bless you sir
$$\mathrm{Ohh},\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *