Question Number 154400 by mathdanisur last updated on 18/Sep/21
$$\mathrm{x}\sqrt{\mathrm{y}}\:-\:\mathrm{y}\sqrt{\mathrm{x}}\:=\:\mathrm{82} \\ $$$$\mathrm{x}\sqrt{\mathrm{x}}\:+\:\mathrm{y}\sqrt{\mathrm{y}}\:=\:\mathrm{83} \\ $$$$\frac{\mathrm{9}}{\mathrm{5}}\:\centerdot\:\left(\mathrm{x}\:+\:\mathrm{y}\right)\:=\:? \\ $$
Answered by TheHoneyCat last updated on 20/Sep/21
![just so you understand my mess (1) will be the first equation and (2) the other other names will be given as I go (in blue) (2) − (1) ⇒ (x+y)(√x)+(y−x)(√y)=1 (3) 82(3)−(1)⇒ (82x+81y)(√x)+(82y−81x)(√y)=0 (1′) 83(3)−(2)⇒ (82x+83y)(√x)+(82y−83x)(√y)=0 or to be more precise : (1)∧(2) ⇔ (1′)∧(2′) its just linear algebra. nothing fancy, its just to be shure I don′t forget a constraint { (((1′))),(((2′))) :}⇔ { (((1′))),(([(2′)−(1′)]÷2)) :}⇔ { (((82x+81y)(√x)+(82y−81x)(√y)=0)),((y(√x)−x(√y)=0 (4))) :} ⇔ { (([(1′)−(4)]÷82)),(((4))) :}⇔ { ((x(√x)+y(√y)=0 (5))),((y(√x)−x(√y)=0)) :} ⇔ ((x,y),(y,(−x)) )× (((√x)),((√y)) )=0 and by the way ((√x),(√y))=(0,0) ⇔ (x,y)=(0,0) is not a solution so ((x,y),(y,(−x)) ) is not invertible ie (4)∧(5)⇒−x^2 −y^2 =0 ⇒ (x,y)=(0,0) so there are no solutions_■ by the way, this could have beenredicted with geogebra indeed, graphicaly (1)⇒x≥40 (ruffly) and (2) ⇒ x≤20 (again, its just an idea) of course you could say there migth be solutions elsewere but if you pay attention the curves discribded by (1) and (2) are continuous since their edges can be observed on geogebra you can be sure you are seeing everything](https://www.tinkutara.com/question/Q154702.png)
$$\mathrm{just}\:\mathrm{so}\:\mathrm{you}\:\mathrm{understand}\:\mathrm{my}\:\mathrm{mess}\:\left(\mathrm{1}\right)\:\mathrm{will}\:\mathrm{be}\:\mathrm{the}\:\mathrm{first}\:\mathrm{equation}\:\mathrm{and}\:\left(\mathrm{2}\right)\:\mathrm{the}\:\mathrm{other} \\ $$$$\mathrm{other}\:\mathrm{names}\:\mathrm{will}\:\mathrm{be}\:\mathrm{given}\:\mathrm{as}\:\mathrm{I}\:\mathrm{go}\:\left({in}\:{blue}\right) \\ $$$$ \\ $$$$\left(\mathrm{2}\right)\:\:−\:\:\left(\mathrm{1}\right)\:\Rightarrow\:\left({x}+{y}\right)\sqrt{{x}}+\left({y}−{x}\right)\sqrt{{y}}=\mathrm{1}\:\left(\mathrm{3}\right) \\ $$$$\mathrm{82}\left(\mathrm{3}\right)−\left(\mathrm{1}\right)\Rightarrow\:\left(\mathrm{82}{x}+\mathrm{81}{y}\right)\sqrt{{x}}+\left(\mathrm{82}{y}−\mathrm{81}{x}\right)\sqrt{{y}}=\mathrm{0}\:\left(\mathrm{1}'\right) \\ $$$$\mathrm{83}\left(\mathrm{3}\right)−\left(\mathrm{2}\right)\Rightarrow\:\left(\mathrm{82}{x}+\mathrm{83}{y}\right)\sqrt{{x}}+\left(\mathrm{82}{y}−\mathrm{83}{x}\right)\sqrt{{y}}=\mathrm{0} \\ $$$$ \\ $$$$\mathrm{or}\:\mathrm{to}\:\mathrm{be}\:\mathrm{more}\:\mathrm{precise}\:: \\ $$$$\left(\mathrm{1}\right)\wedge\left(\mathrm{2}\right)\:\Leftrightarrow\:\:\left(\mathrm{1}'\right)\wedge\left(\mathrm{2}'\right) \\ $$$${its}\:{just}\:{linear}\:{algebra}. \\ $$$${nothing}\:{fancy},\:{its}\:{just}\:{to}\:{be}\:{shure}\:{I}\:{don}'{t}\:{forget}\:{a}\:{constraint} \\ $$$$ \\ $$$$\begin{cases}{\left(\mathrm{1}'\right)}\\{\left(\mathrm{2}'\right)}\end{cases}\Leftrightarrow\begin{cases}{\left(\mathrm{1}'\right)}\\{\left[\left(\mathrm{2}'\right)−\left(\mathrm{1}'\right)\right]\boldsymbol{\div}\mathrm{2}}\end{cases}\Leftrightarrow\begin{cases}{\left(\mathrm{82}{x}+\mathrm{81}{y}\right)\sqrt{{x}}+\left(\mathrm{82}{y}−\mathrm{81}{x}\right)\sqrt{{y}}=\mathrm{0}}\\{{y}\sqrt{{x}}−{x}\sqrt{{y}}=\mathrm{0}\:\left(\mathrm{4}\right)}\end{cases} \\ $$$$\Leftrightarrow\begin{cases}{\left[\left(\mathrm{1}'\right)−\left(\mathrm{4}\right)\right]\boldsymbol{\div}\mathrm{82}}\\{\left(\mathrm{4}\right)}\end{cases}\Leftrightarrow\begin{cases}{{x}\sqrt{{x}}+{y}\sqrt{{y}}=\mathrm{0}\:\left(\mathrm{5}\right)}\\{{y}\sqrt{{x}}−{x}\sqrt{{y}}=\mathrm{0}}\end{cases} \\ $$$$\Leftrightarrow\begin{pmatrix}{{x}}&{{y}}\\{{y}}&{−{x}}\end{pmatrix}×\begin{pmatrix}{\sqrt{{x}}}\\{\sqrt{{y}}}\end{pmatrix}=\mathrm{0}\:{and}\:{by}\:{the}\:{way}\:\left(\sqrt{{x}},\sqrt{{y}}\right)=\left(\mathrm{0},\mathrm{0}\right)\:\Leftrightarrow\:\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right)\:{is}\:{not}\:{a}\:{solution} \\ $$$$\mathrm{so}\:\begin{pmatrix}{{x}}&{{y}}\\{{y}}&{−{x}}\end{pmatrix}\:\mathrm{is}\:\mathrm{not}\:\mathrm{invertible} \\ $$$${ie}\:\left(\mathrm{4}\right)\wedge\left(\mathrm{5}\right)\Rightarrow−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0}\:\Rightarrow\:\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right) \\ $$$$\: \\ $$$$\mathrm{so}\:\mathrm{there}\:\mathrm{are}\:\mathrm{no}\:\mathrm{solutions}_{\blacksquare} \\ $$$$ \\ $$$${by}\:{the}\:{way},\:{this}\:{could}\:{have}\:{beenredicted}\:{with}\:{geogebra} \\ $$$${indeed},\:{graphicaly}\:\left(\mathrm{1}\right)\Rightarrow{x}\geqslant\mathrm{40}\:\left({ruffly}\right) \\ $$$${and}\:\left(\mathrm{2}\right)\:\Rightarrow\:{x}\leqslant\mathrm{20}\:\left({again},\:{its}\:{just}\:{an}\:{idea}\right) \\ $$$${of}\:{course}\:{you}\:{could}\:{say}\:{there}\:{migth}\:{be}\:{solutions}\:{elsewere} \\ $$$${but}\:{if}\:{you}\:{pay}\:{attention}\:{the}\:{curves}\:{discribded}\:{by}\:\left(\mathrm{1}\right)\:{and}\:\left(\mathrm{2}\right)\:{are}\:{continuous} \\ $$$${since}\:{their}\:{edges}\:{can}\:{be}\:{observed}\:{on}\:{geogebra}\:{you}\:{can}\:{be}\:{sure}\:{you}\:{are}\:{seeing}\:{everything} \\ $$$$ \\ $$
Commented by TheHoneyCat last updated on 20/Sep/21
