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Question Number 77028 by TawaTawa last updated on 02/Jan/20

Answered by mind is power last updated on 02/Jan/20

39 x−y −z(x−y)=0 y−z−x(y−z)=0 x^2 +y^2 +z^2 =6 (x−y)(1−z)=0 (y−z)(1−x)=0 if z=1 ⇒1−x=0⇒x=1 y−z=0⇒y=z so all cases ⇒(x=1 or y=z) 40, (1) ,(2),(3) are 1 st 2nd third equation⇒ 2.(1)+1.(2)⇒3x+(2k−1)z=9 7(2)+2(3)⇒(7+2k)x+3z=27 ⇒ ((x),(z) ). (((3 (2k−1))),(((7+2k) 3)) )= ((9),((27)) ) if 9−(2k−1)(7+2k)=0 ⇒9−(12k+4k^2 −7)=0 ⇒k^2 +3k−4=0 (k^2 −(1−4)k+(−4)(1)=0 k∈{1,−4} if k∉{1,−4} we have unique solution ((x),(z) )= ((9),((27)) ).(1/(−4k^2 −12k+16)) (((3 −(2k−1))),((−(7+2k) 3)) ) ((x),(z) )=(1/(−4k^2 −12k+16)) (((54−54k)),((18−18k)) ) ((x),(z) )=((1−k)/(−4(k−1)(k+4))) (((54)),((18)) )= ((((54)/(4(k+4)))),(((18)/(4(k+4))) ) replAce in 1 we get Y if k∈{1,−4} we get 3x+z=9 ⇒z=9−3x parametric solution if k=−4⇒3x−9z=9 −x+3z=27 no solution

$$\mathrm{39} \\ $$$$\mathrm{x}−\mathrm{y}\:−\mathrm{z}\left(\mathrm{x}−\mathrm{y}\right)=\mathrm{0} \\ $$$$\mathrm{y}−\mathrm{z}−\mathrm{x}\left(\mathrm{y}−\mathrm{z}\right)=\mathrm{0} \\ $$$$\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\mathrm{6} \\ $$$$\left(\mathrm{x}−\mathrm{y}\right)\left(\mathrm{1}−\mathrm{z}\right)=\mathrm{0} \\ $$$$\left(\mathrm{y}−\mathrm{z}\right)\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{z}=\mathrm{1} \\ $$$$\Rightarrow\mathrm{1}−\mathrm{x}=\mathrm{0}\Rightarrow\mathrm{x}=\mathrm{1} \\ $$$$\mathrm{y}−\mathrm{z}=\mathrm{0}\Rightarrow\mathrm{y}=\mathrm{z} \\ $$$$\mathrm{so}\:\mathrm{all}\:\mathrm{cases}\:\Rightarrow\left(\mathrm{x}=\mathrm{1}\:\mathrm{or}\:\mathrm{y}=\mathrm{z}\right) \\ $$$$\mathrm{40},\:\left(\mathrm{1}\right)\:\:,\left(\mathrm{2}\right),\left(\mathrm{3}\right)\:\:\mathrm{are}\:\mathrm{1}\:\mathrm{st}\:\mathrm{2nd}\:\mathrm{third}\:\mathrm{equation}\Rightarrow \\ $$$$\mathrm{2}.\left(\mathrm{1}\right)+\mathrm{1}.\left(\mathrm{2}\right)\Rightarrow\mathrm{3x}+\left(\mathrm{2k}−\mathrm{1}\right)\mathrm{z}=\mathrm{9} \\ $$$$\mathrm{7}\left(\mathrm{2}\right)+\mathrm{2}\left(\mathrm{3}\right)\Rightarrow\left(\mathrm{7}+\mathrm{2k}\right)\mathrm{x}+\mathrm{3z}=\mathrm{27} \\ $$$$\Rightarrow\begin{pmatrix}{\mathrm{x}}\\{\mathrm{z}}\end{pmatrix}.\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\left(\mathrm{2k}−\mathrm{1}\right)}\\{\left(\mathrm{7}+\mathrm{2k}\right)\:\:\:\:\mathrm{3}}\end{pmatrix}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{27}}\end{pmatrix} \\ $$$$\mathrm{if}\:\mathrm{9}−\left(\mathrm{2k}−\mathrm{1}\right)\left(\mathrm{7}+\mathrm{2k}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{9}−\left(\mathrm{12k}+\mathrm{4k}^{\mathrm{2}} −\mathrm{7}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{k}^{\mathrm{2}} +\mathrm{3k}−\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{k}^{\mathrm{2}} −\left(\mathrm{1}−\mathrm{4}\right)\mathrm{k}+\left(−\mathrm{4}\right)\left(\mathrm{1}\right)=\mathrm{0}\right. \\ $$$$\mathrm{k}\in\left\{\mathrm{1},−\mathrm{4}\right\} \\ $$$$\mathrm{if}\:\mathrm{k}\notin\left\{\mathrm{1},−\mathrm{4}\right\} \\ $$$$\mathrm{we}\:\mathrm{have}\:\mathrm{unique}\:\mathrm{solution} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{z}}\end{pmatrix}=\begin{pmatrix}{\mathrm{9}}\\{\mathrm{27}}\end{pmatrix}.\frac{\mathrm{1}}{−\mathrm{4k}^{\mathrm{2}} −\mathrm{12k}+\mathrm{16}}\begin{pmatrix}{\mathrm{3}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−\left(\mathrm{2k}−\mathrm{1}\right)}\\{−\left(\mathrm{7}+\mathrm{2k}\right)\:\:\:\:\:\:\mathrm{3}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{z}}\end{pmatrix}=\frac{\mathrm{1}}{−\mathrm{4k}^{\mathrm{2}} −\mathrm{12k}+\mathrm{16}}\begin{pmatrix}{\mathrm{54}−\mathrm{54k}}\\{\mathrm{18}−\mathrm{18k}}\end{pmatrix} \\ $$$$\begin{pmatrix}{\mathrm{x}}\\{\mathrm{z}}\end{pmatrix}=\frac{\mathrm{1}−\mathrm{k}}{−\mathrm{4}\left(\mathrm{k}−\mathrm{1}\right)\left(\mathrm{k}+\mathrm{4}\right)}\begin{pmatrix}{\mathrm{54}}\\{\mathrm{18}}\end{pmatrix}=\begin{pmatrix}{\frac{\mathrm{54}}{\mathrm{4}\left(\mathrm{k}+\mathrm{4}\right)}}\\{\frac{\mathrm{18}}{\mathrm{4}\left(\mathrm{k}+\mathrm{4}\right.}}\end{pmatrix} \\ $$$$\mathrm{replAce}\:\:\mathrm{in}\:\mathrm{1}\:\mathrm{we}\:\mathrm{get}\:\mathrm{Y} \\ $$$$\mathrm{if}\:\mathrm{k}\in\left\{\mathrm{1},−\mathrm{4}\right\} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{3x}+\mathrm{z}=\mathrm{9} \\ $$$$\Rightarrow\mathrm{z}=\mathrm{9}−\mathrm{3x} \\ $$$$\mathrm{parametric}\:\mathrm{solution} \\ $$$$\mathrm{if}\:\mathrm{k}=−\mathrm{4}\Rightarrow\mathrm{3x}−\mathrm{9z}=\mathrm{9} \\ $$$$−\mathrm{x}+\mathrm{3z}=\mathrm{27} \\ $$$$\mathrm{no}\:\mathrm{solution} \\ $$$$ \\ $$$$ \\ $$

Commented by behi83417@gmail.com last updated on 02/Jan/20

wellllllllllcom back sir:mind is power. being of you is nessesary in this forum. Q#76954,5,6 are waiting for your attention.

$$\mathrm{wellllllllllcom}\:\mathrm{back}\:\mathrm{sir}:\mathrm{mind}\:\mathrm{is}\:\mathrm{power}. \\ $$$$\mathrm{being}\:\mathrm{of}\:\mathrm{you}\:\mathrm{is}\:\mathrm{nessesary}\:\mathrm{in}\:\mathrm{this}\:\mathrm{forum}. \\ $$$$\mathrm{Q}#\mathrm{76954},\mathrm{5},\mathrm{6}\:\mathrm{are}\:\mathrm{waiting}\:\mathrm{for}\:\mathrm{your}\:\mathrm{attention}. \\ $$

Commented by TawaTawa last updated on 02/Jan/20

God bless you sir

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

Commented by mind is power last updated on 02/Jan/20

Thanx sir Tawa Tawa

$$\mathrm{Thanx}\:\mathrm{sir}\:\mathrm{Tawa}\:\mathrm{Tawa} \\ $$$$ \\ $$

Commented by mind is power last updated on 02/Jan/20

I will chek it sir Tomorrow sam hour that now will post solution Sir , i will have mor Times in 2 days

$$\mathrm{I}\:\mathrm{will}\:\mathrm{chek}\:\mathrm{it}\:\mathrm{sir}\:\mathrm{Tomorrow}\:\mathrm{sam}\:\mathrm{hour}\:\mathrm{that}\:\mathrm{now} \\ $$$$\:\mathrm{will}\:\mathrm{post}\:\mathrm{solution}\:\mathrm{Sir}\:,\:\mathrm{i}\:\mathrm{will}\:\mathrm{have}\:\mathrm{mor}\:\mathrm{Times}\:\mathrm{in}\:\mathrm{2}\:\mathrm{days} \\ $$

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