Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

$$\mathrm{Thankyou}\:\mathrm{Ser} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com