Question Number 4430 by alib last updated on 24/Jan/16
$$\left\{\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \right. \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{15}{a}^{\mathrm{2}} \\ $$$$ \\ $$$${Solve}\:{the}\:{system}\:{of}\:{equation} \\ $$$$ \\ $$
Answered by Rasheed Soomro last updated on 24/Jan/16
![{(x−y)(x^2 −y^2 )=3a^2 ..........................I (x+y)(x^2 +y^2 )=15a^2 ..........................II (((x+y)(x^2 +y^2 ))/({(x−y)(x^2 −y^2 )))=((15a^2 )/(3a^2 )) [((II)/I)] (((x+y)(x^2 +y^2 ))/({(x+y)(x−y)^2 ))=5 x^2 +y^2 =5(x−y)^2 x^2 +y^2 =5x^2 −10xy+5y^2 4x^2 −10xy+4y^2 =0 2x^2 −5xy+2y^2 =0 2x^2 −4xy−xy+2y^2 =0 2x(x−2y)−y(x−2y)=0 (x−2y)(2x−y)=0 y=(x/2) ∣ y=2x When y=(x/2) , I will be (x−(x/2))(x^2 −((x/2))^2 )=3a^2 (x/2)(((3x^2 )/4))=3a^2 (x^3 /8)=a^2 x^3 =8a^2 x^3 −8a^2 =0 (x)^3 −(2a^(2/3) )^3 =0 (x−2a^(2/3) )(x^2 +2xa^(2/3) +4a^(4/3) )=0 x=2a^(2/3) ∣ x=2a^(2/3) ω ∣ x=2a^(2/3) ω^2 .....∗ y=(x/2)⇒ y=a^(2/3) ∣ y=a^(2/3) ω ∣ y=a^(2/3) ω^2 When y=2x, I will be (x−2x)(x^2 −(2x)^2 )=3a^2 (-x)(-3x^2 )=3a^2 x^3 −a^2 =0 (x)^3 −(a^(2/3) )^3 =0 (x−a^(2/3) )(x^2 +xa^(2/3) +a^(4/3) )=0 x=a^(2/3) ∣ x=a^(2/3) ω ∣ x=a^(2/3) ω^2 ........∗ y=2x⇒ y=2a^(2/3) ∣ y=2a^(2/3) ω ∣ y=2a^(2/3) ω^2 Solution Set: { (a^(2/3) ,2a^(2/3) ),(a^(2/3) ω , 2a^(2/3) ω),(a^(2/3) ω^2 , 2a^(2/3) ω^2 ), (2a^(2/3) , a^(2/3) ),(2a^(2/3) ω , a^(2/3) ω),(2a^(2/3) ω^2 , a^(2/3) ω^2 ) }..∗ ∗ ω is cuberoot of unity.](https://www.tinkutara.com/question/Q4432.png)
$$\left\{\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} ……………………..\mathrm{I}\right. \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{15}{a}^{\mathrm{2}} ……………………..\mathrm{II} \\ $$$$\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\left\{\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)\right.}=\frac{\mathrm{15}{a}^{\mathrm{2}} }{\mathrm{3}{a}^{\mathrm{2}} }\:\:\:\:\:\:\left[\frac{\mathrm{II}}{\mathrm{I}}\right] \\ $$$$ \\ $$$$\frac{\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}{\left\{\left({x}+{y}\right)\left({x}−{y}\right)^{\mathrm{2}} \right.}=\mathrm{5} \\ $$$$ \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}\left({x}−{y}\right)^{\mathrm{2}} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{5}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{5}{y}^{\mathrm{2}} \\ $$$$\mathrm{4}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{4}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{xy}−{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{2}{x}\left({x}−\mathrm{2}{y}\right)−{y}\left({x}−\mathrm{2}{y}\right)=\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{y}\right)\left(\mathrm{2}{x}−{y}\right)=\mathrm{0} \\ $$$${y}=\frac{{x}}{\mathrm{2}}\:\mid\:{y}=\mathrm{2}{x} \\ $$$${When}\:{y}=\frac{{x}}{\mathrm{2}}\:,\:\mathrm{I}\:{will}\:{be} \\ $$$$\left({x}−\frac{{x}}{\mathrm{2}}\right)\left({x}^{\mathrm{2}} −\left(\frac{{x}}{\mathrm{2}}\right)^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\frac{{x}}{\mathrm{2}}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\mathrm{4}}\right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\frac{{x}^{\mathrm{3}} }{\mathrm{8}}={a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} =\mathrm{8}{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\mathrm{8}{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}\right)^{\mathrm{3}} −\left(\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({x}−\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \right)\left({x}^{\mathrm{2}} +\mathrm{2}{xa}^{\mathrm{2}/\mathrm{3}} +\mathrm{4}{a}^{\mathrm{4}/\mathrm{3}} \right)=\mathrm{0} \\ $$$${x}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\:\mid\:{x}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\omega\:\:\mid\:{x}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\omega^{\mathrm{2}} …..\ast \\ $$$${y}=\frac{{x}}{\mathrm{2}}\Rightarrow\:{y}={a}^{\mathrm{2}/\mathrm{3}} \:\mid\:{y}={a}^{\mathrm{2}/\mathrm{3}} \omega\:\mid\:{y}={a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} \\ $$$${When}\:{y}=\mathrm{2}{x},\:\mathrm{I}\:{will}\:{be} \\ $$$$\left({x}−\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} −\left(\mathrm{2}{x}\right)^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\left(-{x}\right)\left(-\mathrm{3}{x}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({x}\right)^{\mathrm{3}} −\left({a}^{\mathrm{2}/\mathrm{3}} \right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\left({x}−{a}^{\mathrm{2}/\mathrm{3}} \right)\left({x}^{\mathrm{2}} +{xa}^{\mathrm{2}/\mathrm{3}} +{a}^{\mathrm{4}/\mathrm{3}} \right)=\mathrm{0} \\ $$$${x}={a}^{\mathrm{2}/\mathrm{3}} \:\mid\:{x}={a}^{\mathrm{2}/\mathrm{3}} \omega\:\mid\:{x}={a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} ……..\ast \\ $$$${y}=\mathrm{2}{x}\Rightarrow\:{y}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\:\mid\:{y}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\omega\:\:\mid\:{y}=\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \:\omega^{\mathrm{2}} \\ $$$$ \\ $$$${Solution}\:{Set}: \\ $$$$\left\{\:\left({a}^{\mathrm{2}/\mathrm{3}} ,\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \right),\left({a}^{\mathrm{2}/\mathrm{3}} \omega\:,\:\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \omega\right),\left({a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} \:,\:\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} \right),\right. \\ $$$$\left.\:\:\:\:\left(\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} ,\:{a}^{\mathrm{2}/\mathrm{3}} \right),\left(\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \omega\:,\:{a}^{\mathrm{2}/\mathrm{3}} \omega\right),\left(\mathrm{2}{a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} ,\:{a}^{\mathrm{2}/\mathrm{3}} \omega^{\mathrm{2}} \right)\:\:\right\}..\ast \\ $$$$\ast\:\omega\:{is}\:{cuberoot}\:{of}\:{unity}. \\ $$
Commented by Yozzii last updated on 24/Jan/16
$${noice} \\ $$
Answered by Rasheed Soomro last updated on 24/Jan/16
![An Other way (x−y)(x^2 −y^2 )=3a^2 ...................I (x+y)(x^2 +y^2 )=15a^2 .................II 5(x+y)(x−y)^2 =15a^2 ...............III [I×5] 5(x+y)(x−y)^2 −(x+y)(x^2 +y^2 )=0 [III − II] (x+y){5(x−y)^2 −(x^2 +y^2 )}=0 x+y=0 ∣ 5x^2 −10xy+5y^2 −x^2 −y^2 =0 y=−x ∣ 4x^2 −10xy+4y^2 =0⇒2x^2 −5xy+2y^2 =0 When y=−x, I will be (x−(-x))(x^2 −(-x)^2 )=3a^2 2x(x^2 −x^2 )=3a^2 0=3a^2 So for y=−x the system has no solution. For the solution of 2x^2 −5xy+2y^2 =0 please see my other answer.](https://www.tinkutara.com/question/Q4434.png)
$$\mathrm{An}\:\mathrm{Other}\:\mathrm{way} \\ $$$$\left({x}−{y}\right)\left({x}^{\mathrm{2}} −{y}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} ……………….\mathrm{I} \\ $$$$\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{15}{a}^{\mathrm{2}} ……………..\mathrm{II} \\ $$$$\mathrm{5}\left({x}+{y}\right)\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{15}{a}^{\mathrm{2}} ……………\mathrm{III}\:\left[\mathrm{I}×\mathrm{5}\right] \\ $$$$\mathrm{5}\left({x}+{y}\right)\left({x}−{y}\right)^{\mathrm{2}} −\left({x}+{y}\right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{0}\:\:\:\left[\mathrm{III}\:−\:\mathrm{II}\right] \\ $$$$\left({x}+{y}\right)\left\{\mathrm{5}\left({x}−{y}\right)^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\right\}=\mathrm{0} \\ $$$${x}+{y}=\mathrm{0}\:\:\mid\:\mathrm{5}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{5}{y}^{\mathrm{2}} −{x}^{\mathrm{2}} −{y}^{\mathrm{2}} =\mathrm{0} \\ $$$${y}=−{x}\:\:\mid\:\mathrm{4}{x}^{\mathrm{2}} −\mathrm{10}{xy}+\mathrm{4}{y}^{\mathrm{2}} =\mathrm{0}\Rightarrow\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{When}\:{y}=−{x},\:\mathrm{I}\:\:{will}\:{be} \\ $$$$\left({x}−\left(-{x}\right)\right)\left({x}^{\mathrm{2}} −\left(-{x}\right)^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\mathrm{2}{x}\left({x}^{\mathrm{2}} −{x}^{\mathrm{2}} \right)=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\mathrm{0}=\mathrm{3}{a}^{\mathrm{2}} \:\:{So}\:{for}\:{y}=−{x}\:{the}\:{system}\:{has}\:{no}\:{solution}. \\ $$$${For}\:{the}\:{solution}\:{of}\:\mathrm{2}{x}^{\mathrm{2}} −\mathrm{5}{xy}+\mathrm{2}{y}^{\mathrm{2}} =\mathrm{0}\:{please} \\ $$$${see}\:{my}\:{other}\:{answer}. \\ $$