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x-2-x-6y-y-2-y-6x-x-y-




Question Number 207092 by hardmath last updated on 06/May/24
 { ((x^2  = x − 6y)),((y^2  = y − 6x)) :}     ⇒    x + y = ?
$$\begin{cases}{\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{x}\:−\:\mathrm{6y}}\\{\mathrm{y}^{\mathrm{2}} \:=\:\mathrm{y}\:−\:\mathrm{6x}}\end{cases}\:\:\:\:\:\Rightarrow\:\:\:\:\mathrm{x}\:+\:\mathrm{y}\:=\:? \\ $$
Answered by A5T last updated on 06/May/24
x^2 −y^2 =x−y−6y+6x=7x−7y  ⇒(x−y)(x+y)=7(x−y)  x−y≠0 ⇒x+y=7  when x=y; x^2 =y^2 =−5x⇒x=y=0 or x=y=−5  ⇒x+y=0 or x+y=−10  ⇒x+y∈{−10,0,7}
$${x}^{\mathrm{2}} −{y}^{\mathrm{2}} ={x}−{y}−\mathrm{6}{y}+\mathrm{6}{x}=\mathrm{7}{x}−\mathrm{7}{y} \\ $$$$\Rightarrow\left({x}−{y}\right)\left({x}+{y}\right)=\mathrm{7}\left({x}−{y}\right) \\ $$$${x}−{y}\neq\mathrm{0}\:\Rightarrow{x}+{y}=\mathrm{7} \\ $$$${when}\:{x}={y};\:{x}^{\mathrm{2}} ={y}^{\mathrm{2}} =−\mathrm{5}{x}\Rightarrow{x}={y}=\mathrm{0}\:{or}\:{x}={y}=−\mathrm{5} \\ $$$$\Rightarrow{x}+{y}=\mathrm{0}\:{or}\:{x}+{y}=−\mathrm{10} \\ $$$$\Rightarrow{x}+{y}\in\left\{−\mathrm{10},\mathrm{0},\mathrm{7}\right\} \\ $$
Answered by mr W last updated on 06/May/24
(i)+(ii):  x^2 +y^2 =−5(x+y) ⇒x+y≤0    (i)−(ii):  (x+y)(x−y)=7(x−y)  x≠y: ⇒x+y=7>0 ⇒rejected  x=y: x^2 =−5x ⇒x=0 or −5   ⇒x+y=0 or −10
$$\left({i}\right)+\left({ii}\right): \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =−\mathrm{5}\left({x}+{y}\right)\:\Rightarrow{x}+{y}\leqslant\mathrm{0} \\ $$$$ \\ $$$$\left({i}\right)−\left({ii}\right): \\ $$$$\left({x}+{y}\right)\left({x}−{y}\right)=\mathrm{7}\left({x}−{y}\right) \\ $$$${x}\neq{y}:\:\Rightarrow{x}+{y}=\mathrm{7}>\mathrm{0}\:\Rightarrow{rejected} \\ $$$${x}={y}:\:{x}^{\mathrm{2}} =−\mathrm{5}{x}\:\Rightarrow{x}=\mathrm{0}\:{or}\:−\mathrm{5}\: \\ $$$$\Rightarrow{x}+{y}=\mathrm{0}\:{or}\:−\mathrm{10} \\ $$
Commented by Frix last updated on 06/May/24
But the solutions of the system are  x=y=0 ⇒ x+y=0  x=y=−5 ⇒ x+y=−10  x=(7/2)±((√(119))/2)i∧y=conj x ⇒ x+y=7
$$\mathrm{But}\:\mathrm{the}\:\mathrm{solutions}\:\mathrm{of}\:\mathrm{the}\:\mathrm{system}\:\mathrm{are} \\ $$$${x}={y}=\mathrm{0}\:\Rightarrow\:{x}+{y}=\mathrm{0} \\ $$$${x}={y}=−\mathrm{5}\:\Rightarrow\:{x}+{y}=−\mathrm{10} \\ $$$${x}=\frac{\mathrm{7}}{\mathrm{2}}\pm\frac{\sqrt{\mathrm{119}}}{\mathrm{2}}\mathrm{i}\wedge{y}=\mathrm{conj}\:{x}\:\Rightarrow\:{x}+{y}=\mathrm{7} \\ $$
Commented by mr W last updated on 06/May/24
i live in R world :)
$$\left.{i}\:{live}\:{in}\:\mathbb{R}\:{world}\::\right) \\ $$
Commented by Frix last updated on 06/May/24
...but the real world is complex... ��
Commented by mr W last updated on 06/May/24
indeed! the real world is too  complex to be real to me...
$${indeed}!\:{the}\:{real}\:{world}\:{is}\:{too} \\ $$$${complex}\:{to}\:{be}\:{real}\:{to}\:{me}… \\ $$

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