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 139874 by bramlexs22 last updated on 02/May/21

{ ((xy+24 = (x^3 /y))),((xy−6 = (y^3 /x))) :}

$$\begin{cases}{\mathrm{xy}+\mathrm{24}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}}\\{\mathrm{xy}−\mathrm{6}\:=\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}}}\end{cases} \\ $$

Answered by EDWIN88 last updated on 03/May/21

(1)xy +24 = (x^3 /y) (2) xy−6 = (y^3 /x) (1)×(2)⇒(xy+24)(xy−6)=(x^(3^2 ) /y).(y^(3^2 ) /x) ⇒(xy+24)(xy−6) = (xy)^2 let xy = u ⇒(u+24)(u−6)=u^2 ⇒^3 18u = 24.6 ; u=8 →y=(8/x) input to eq (2)⇒ 8−6 = (8^3 /x^4 ) ; 2x^4 = 8^3 ⇒x^4 =4×4^3 ; (x^2 −4^2 )(x^2 +4^2 )=0 ⇒(x−4)(x+4)(x^2 +16)=0 ⇒ { ((x=4⇒y=2)),((x=−4⇒y=−2)),((x=±4i⇒y=±2i)) :} for x,y∈C

$$\left(\mathrm{1}\right)\mathrm{xy}\:+\mathrm{24}\:=\:\frac{\mathrm{x}^{\mathrm{3}} }{\mathrm{y}}\:\:\:\:\:\:\left(\mathrm{2}\right)\:\mathrm{xy}−\mathrm{6}\:=\:\frac{\mathrm{y}^{\mathrm{3}} }{\mathrm{x}} \\ $$$$\left(\mathrm{1}\right)×\left(\mathrm{2}\right)\Rightarrow\left(\mathrm{xy}+\mathrm{24}\right)\left(\mathrm{xy}−\mathrm{6}\right)=\frac{\mathrm{x}^{\cancel{\mathrm{3}}\:^{\mathrm{2}} } }{\cancel{\mathrm{y}}}.\frac{\mathrm{y}^{\cancel{\mathrm{3}}\:^{\mathrm{2}} } }{\cancel{\mathrm{x}}} \\ $$$$\Rightarrow\left(\mathrm{xy}+\mathrm{24}\right)\left(\mathrm{xy}−\mathrm{6}\right)\:=\:\left(\mathrm{xy}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{xy}\:=\:\mathrm{u}\:\Rightarrow\left(\mathrm{u}+\mathrm{24}\right)\left(\mathrm{u}−\mathrm{6}\right)=\mathrm{u}^{\mathrm{2}} \\ $$$$\Rightarrow^{\mathrm{3}} \cancel{\mathrm{18}u}\:=\:\mathrm{24}.\cancel{\mathrm{6}}\:;\:\mathrm{u}=\mathrm{8}\:\rightarrow\mathrm{y}=\frac{\mathrm{8}}{\mathrm{x}} \\ $$$$\mathrm{input}\:\mathrm{to}\:\mathrm{eq}\:\left(\mathrm{2}\right)\Rightarrow\:\mathrm{8}−\mathrm{6}\:=\:\frac{\mathrm{8}^{\mathrm{3}} }{\mathrm{x}^{\mathrm{4}} }\:;\:\mathrm{2x}^{\mathrm{4}} \:=\:\mathrm{8}^{\mathrm{3}} \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{4}} =\mathrm{4}×\mathrm{4}^{\mathrm{3}} \:;\:\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}^{\mathrm{2}} \right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{x}−\mathrm{4}\right)\left(\mathrm{x}+\mathrm{4}\right)\left(\mathrm{x}^{\mathrm{2}} +\mathrm{16}\right)=\mathrm{0} \\ $$$$\Rightarrow\begin{cases}{\mathrm{x}=\mathrm{4}\Rightarrow\mathrm{y}=\mathrm{2}}\\{\mathrm{x}=−\mathrm{4}\Rightarrow\mathrm{y}=−\mathrm{2}}\\{\mathrm{x}=\pm\mathrm{4i}\Rightarrow\mathrm{y}=\pm\mathrm{2i}}\end{cases} \\ $$$$\mathrm{for}\:\mathrm{x},\mathrm{y}\in\mathrm{C} \\ $$

Answered by MJS_new last updated on 02/May/21

let y=px px^2 +24=(x^2 /p) ⇔ x^2 =((24p)/((1−p^2 ))) px^2 −6=p^3 x^2 ⇔ x^2 =(6/(p(1−p^2 ))) ((24p)/((1−p^2 )))=(6/(p(1−p^2 ))) p^2 =(1/4) ⇒ p=±(1/2) testing all possible pairs x, y leads to x_1 =−4 ∧ y_1 =−2 x_2 =+4 ∧ y_2 =+2 x_3 =−4i ∧ y_3 =+2i x_4 =+4i ∧ y_4 =−2i

$$\mathrm{let}\:{y}={px} \\ $$$${px}^{\mathrm{2}} +\mathrm{24}=\frac{{x}^{\mathrm{2}} }{{p}}\:\Leftrightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{24}{p}}{\left(\mathrm{1}−{p}^{\mathrm{2}} \right)} \\ $$$${px}^{\mathrm{2}} −\mathrm{6}={p}^{\mathrm{3}} {x}^{\mathrm{2}} \:\Leftrightarrow\:{x}^{\mathrm{2}} =\frac{\mathrm{6}}{{p}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)} \\ $$$$\frac{\mathrm{24}{p}}{\left(\mathrm{1}−{p}^{\mathrm{2}} \right)}=\frac{\mathrm{6}}{{p}\left(\mathrm{1}−{p}^{\mathrm{2}} \right)} \\ $$$${p}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\:\Rightarrow\:{p}=\pm\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{testing}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{pairs}\:{x},\:{y}\:\mathrm{leads}\:\mathrm{to} \\ $$$${x}_{\mathrm{1}} =−\mathrm{4}\:\wedge\:{y}_{\mathrm{1}} =−\mathrm{2} \\ $$$${x}_{\mathrm{2}} =+\mathrm{4}\:\wedge\:{y}_{\mathrm{2}} =+\mathrm{2} \\ $$$${x}_{\mathrm{3}} =−\mathrm{4i}\:\wedge\:{y}_{\mathrm{3}} =+\mathrm{2i} \\ $$$${x}_{\mathrm{4}} =+\mathrm{4i}\:\wedge\:{y}_{\mathrm{4}} =−\mathrm{2i} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com