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Find-x-y-x-Q-and-y-Z-such-that-2020-x-2-y-2-2019-x-y-2021xy-

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Question Number 150531 by mathdanisur last updated on 13/Aug/21
Find x;y ; x∈Q and y∈Z such that: 2020(x^2 + y^2 ) + 2019(x + y) = 2021xy
$$\mathrm{Find}\:\:\mathrm{x};\mathrm{y}\:\:;\:\:\mathrm{x}\in\mathrm{Q}\:\:\mathrm{and}\:\:\mathrm{y}\in\mathrm{Z}\:\:\mathrm{such}\:\mathrm{that}: \\ $$$$\mathrm{2020}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\:+\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{2021xy} \\ $$
Commented by Rasheed.Sindhi last updated on 14/Aug/21
⋘Better than nothing answer⋙_(⌣) ^(⌢) 2020(x^2 + y^2 ) + 2019(x + y) = 2021xy 2020(x^2 + y^2 +2xy−2xy) + 2019(x + y) = 2021xy 2020(x+y)^2 + 2019(x + y) − 6061xy=0 △=2019^2 −4(2020)(−6061xy) △=2019^2 +4(2020)(6061xy) For xy=4, △ is perfect square So let xy=4 x+y=((−2019±14141)/(4040))=((6061)/(2020)),−4 (xy,x+y)=(4,((6061)/(2020))),(4,−4) (x+y)^2 −4xy=(x−y)^2 (((6061)/(2020)))^2 −4(4)=(x−y)^2 x−y is not real (−4)^2 −4(4)=(x−y)^2 =0 x−y=0 x+y=−4 x=−2,y=−2
$$\underset{\smile} {\overset{\frown} {\lll\mathcal{B}{etter}\:{than}\:{nothing}\:{answer}\ggg}} \\ $$$$\mathrm{2020}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\:+\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{2021xy} \\ $$$$\mathrm{2020}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} +\mathrm{2}{xy}−\mathrm{2}{xy}\right)\:+\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{2021xy} \\ $$$$\mathrm{2020}\left({x}+{y}\right)^{\mathrm{2}} +\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:−\:\mathrm{6061xy}=\mathrm{0} \\ $$$$\bigtriangleup=\mathrm{2019}^{\mathrm{2}} −\mathrm{4}\left(\mathrm{2020}\right)\left(−\mathrm{6061}{xy}\right) \\ $$$$\bigtriangleup=\mathrm{2019}^{\mathrm{2}} +\mathrm{4}\left(\mathrm{2020}\right)\left(\mathrm{6061}{xy}\right) \\ $$$$\:\:\:\:\:{For}\:{xy}=\mathrm{4},\:\bigtriangleup\:{is}\:{perfect}\:{square} \\ $$$${So}\:{let}\:{xy}=\mathrm{4} \\ $$$${x}+{y}=\frac{−\mathrm{2019}\pm\mathrm{14141}}{\mathrm{4040}}=\frac{\mathrm{6061}}{\mathrm{2020}},−\mathrm{4} \\ $$$$\left({xy},{x}+{y}\right)=\left(\mathrm{4},\frac{\mathrm{6061}}{\mathrm{2020}}\right),\left(\mathrm{4},−\mathrm{4}\right) \\ $$$$\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{4}{xy}=\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\left(\frac{\mathrm{6061}}{\mathrm{2020}}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{4}\right)=\left({x}−{y}\right)^{\mathrm{2}} \\ $$$$\:\:\:{x}−{y}\:{is}\:{not}\:{real} \\ $$$$\left(−\mathrm{4}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{4}\right)=\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$${x}−{y}=\mathrm{0} \\ $$$${x}+{y}=−\mathrm{4} \\ $$$${x}=−\mathrm{2},{y}=−\mathrm{2} \\ $$
Commented by mathdanisur last updated on 13/Aug/21
Thank you Ser, but ans: (0;0)
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser},\:\mathrm{but}\:\mathrm{ans}:\:\left(\mathrm{0};\mathrm{0}\right) \\ $$
Commented by Rasheed.Sindhi last updated on 13/Aug/21
2020(x^2 + y^2 ) + 2019(x + y) = 2021xy 2020( (−2)^2 +(−2)^2 )+2019(−2+(−2) ) =2021(−2)(−2) 2020(8)+2019(−4)=8084 8084=8084 So (−2,−2) also satisfy the given equation.
$$\mathrm{2020}\left(\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \right)\:+\:\mathrm{2019}\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:\mathrm{2021xy} \\ $$$$\mathrm{2020}\left(\:\left(−\mathrm{2}\right)^{\mathrm{2}} +\left(−\mathrm{2}\right)^{\mathrm{2}} \:\right)+\mathrm{2019}\left(−\mathrm{2}+\left(−\mathrm{2}\right)\:\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2021}\left(−\mathrm{2}\right)\left(−\mathrm{2}\right) \\ $$$$\mathrm{2020}\left(\mathrm{8}\right)+\mathrm{2019}\left(−\mathrm{4}\right)=\mathrm{8084} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{8084}=\mathrm{8084} \\ $$$${So}\:\left(−\mathrm{2},−\mathrm{2}\right)\:{also}\:{satisfy}\:{the}\:{given} \\ $$$${equation}. \\ $$
Commented by mathdanisur last updated on 13/Aug/21
Dear Ser, equation has 4 solution
$$\mathrm{Dear}\:\mathrm{Ser},\:\mathrm{equation}\:\mathrm{has}\:\mathrm{4}\:\mathrm{solution} \\ $$
Commented by Rasheed.Sindhi last updated on 13/Aug/21
I didn′t say there is only this solution, I said this is also a solution.Thanks.
$${I}\:{didn}'{t}\:{say}\:{there}\:{is}\:{only}\:{this}\:{solution}, \\ $$$${I}\:{said}\:{this}\:{is}\:{also}\:{a}\:{solution}.{Thanks}. \\ $$$$ \\ $$
Commented by mathdanisur last updated on 14/Aug/21
Thankyou Ser
$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

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