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Question Number 58195 by salaw2000 last updated on 19/Apr/19
(1/x)+(1/y)=(3/4) (x^2 /y)+(y^2 /x)=9 find the value of x and y
$$\frac{\mathrm{1}}{\mathrm{x}}+\frac{\mathrm{1}}{\mathrm{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{y}}+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{x}}=\mathrm{9} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{x}\:\mathrm{and}\:\mathrm{y} \\ $$
Answered by Kunal12588 last updated on 19/Apr/19
(1/x)+(1/y)=(3/4) ⇒((x+y)/(xy))=(3/4) (1) ⇒x+y=((3xy)/4) (2) (x^2 /y)+(y^2 /x)=9 ⇒((x+y)/(xy))(x^2 −xy+y^2 )=9 from (1) ⇒x^2 −xy+y^2 =12 ⇒(x+y)^2 −3xy=12 (3) ⇒(x−y)^2 +xy=12 (4) from (3) x+y=±(√(3xy+12)) but from (2) ((3xy)/4)=(√(3xy+12)) ⇒3∙3x^2 y^2 =16[3(xy+4)] ⇒3x^2 y^2 −16xy−64=0 ⇒xy=((16±(√(16^2 −4∙3∙(−64))))/(2∙3)) ⇒xy=((48)/6),((−16)/6) ⇒xy=8,((−8)/3) (i) xy=8 putting xy in (2) x+y=6 (5) putting xy in (4) (x−y)^2 =12−8 ⇒x−y=±(√4) ⇒x−y=2 or x−y=−2 (a) taking x−y=2 and comparing with (5) 1^(st) answer x=4,y=2 (b) taking x−y=−2 and comparing with (5) 2^(nd) answer x=2,y=4 (ii) xy=((−8)/3) putting xy in (2) x+y=−2 (6) putting xy in (4) (x−y)^2 +xy=12 3(x−y)^2 −8=36 x−y=±(√((44)/3)) (a) taking x−y=(√((44)/3)) and comparing with (6) 3^(rd) answer x=−1+(√((11)/3)),y=−1−(√((11)/3)) (b) taking x−y=−(√((44)/3)) and comparing with (6) 4^(th) answer x=−1−(√((11)/3)),y=−1+(√((11)/3))
$$\frac{\mathrm{1}}{{x}}+\frac{\mathrm{1}}{{y}}=\frac{\mathrm{3}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{{x}+{y}}{{xy}}=\frac{\mathrm{3}}{\mathrm{4}}\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow{x}+{y}=\frac{\mathrm{3}{xy}}{\mathrm{4}}\:\:\:\:\:\:\left(\mathrm{2}\right) \\ $$$$\frac{{x}^{\mathrm{2}} }{{y}}+\frac{{y}^{\mathrm{2}} }{{x}}=\mathrm{9} \\ $$$$\Rightarrow\frac{{x}+{y}}{{xy}}\left({x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} \right)=\mathrm{9} \\ $$$${from}\:\left(\mathrm{1}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{xy}+{y}^{\mathrm{2}} =\mathrm{12} \\ $$$$\Rightarrow\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{3}{xy}=\mathrm{12}\:\:\:\:\:\:\:\:\left(\mathrm{3}\right) \\ $$$$\Rightarrow\left({x}−{y}\right)^{\mathrm{2}} +{xy}=\mathrm{12}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{4}\right) \\ $$$${from}\:\left(\mathrm{3}\right) \\ $$$${x}+{y}=\pm\sqrt{\mathrm{3}{xy}+\mathrm{12}}\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$$${but}\:{from}\:\left(\mathrm{2}\right) \\ $$$$\frac{\mathrm{3}{xy}}{\mathrm{4}}=\sqrt{\mathrm{3}{xy}+\mathrm{12}} \\ $$$$\Rightarrow\mathrm{3}\centerdot\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} =\mathrm{16}\left[\mathrm{3}\left({xy}+\mathrm{4}\right)\right] \\ $$$$\Rightarrow\mathrm{3}{x}^{\mathrm{2}} {y}^{\mathrm{2}} −\mathrm{16}{xy}−\mathrm{64}=\mathrm{0} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{16}\pm\sqrt{\mathrm{16}^{\mathrm{2}} −\mathrm{4}\centerdot\mathrm{3}\centerdot\left(−\mathrm{64}\right)}}{\mathrm{2}\centerdot\mathrm{3}} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{48}}{\mathrm{6}},\frac{−\mathrm{16}}{\mathrm{6}} \\ $$$$\Rightarrow{xy}=\mathrm{8},\frac{−\mathrm{8}}{\mathrm{3}} \\ $$$$\left({i}\right)\:{xy}=\mathrm{8} \\ $$$${putting}\:{xy}\:{in}\:\left(\mathrm{2}\right) \\ $$$${x}+{y}=\mathrm{6}\:\:\:\:\left(\mathrm{5}\right) \\ $$$${putting}\:{xy}\:{in}\:\left(\mathrm{4}\right) \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{12}−\mathrm{8} \\ $$$$\Rightarrow{x}−{y}=\pm\sqrt{\mathrm{4}} \\ $$$$\Rightarrow{x}−{y}=\mathrm{2}\:{or}\:{x}−{y}=−\mathrm{2} \\ $$$$\left({a}\right)\:\:{taking}\:{x}−{y}=\mathrm{2} \\ $$$${and}\:{comparing}\:{with}\:\left(\mathrm{5}\right) \\ $$$$\mathrm{1}^{{st}} \:{answer}\:\:\:{x}=\mathrm{4},{y}=\mathrm{2} \\ $$$$\left({b}\right)\:\:{taking}\:{x}−{y}=−\mathrm{2} \\ $$$${and}\:{comparing}\:{with}\:\left(\mathrm{5}\right) \\ $$$$\mathrm{2}^{{nd}} \:{answer}\:{x}=\mathrm{2},{y}=\mathrm{4} \\ $$$$\left({ii}\right)\:{xy}=\frac{−\mathrm{8}}{\mathrm{3}} \\ $$$${putting}\:{xy}\:{in}\:\left(\mathrm{2}\right) \\ $$$${x}+{y}=−\mathrm{2}\:\:\:\:\:\left(\mathrm{6}\right) \\ $$$${putting}\:{xy}\:{in}\:\left(\mathrm{4}\right) \\ $$$$\left({x}−{y}\right)^{\mathrm{2}} +{xy}=\mathrm{12} \\ $$$$\mathrm{3}\left({x}−{y}\right)^{\mathrm{2}} −\mathrm{8}=\mathrm{36} \\ $$$${x}−{y}=\pm\sqrt{\frac{\mathrm{44}}{\mathrm{3}}} \\ $$$$\left({a}\right)\:\:{taking}\:{x}−{y}=\sqrt{\frac{\mathrm{44}}{\mathrm{3}}} \\ $$$${and}\:{comparing}\:{with}\:\left(\mathrm{6}\right) \\ $$$$\mathrm{3}^{{rd}} \:{answer}\:{x}=−\mathrm{1}+\sqrt{\frac{\mathrm{11}}{\mathrm{3}}},{y}=−\mathrm{1}−\sqrt{\frac{\mathrm{11}}{\mathrm{3}}} \\ $$$$\left({b}\right)\:\:{taking}\:{x}−{y}=−\sqrt{\frac{\mathrm{44}}{\mathrm{3}}} \\ $$$${and}\:{comparing}\:{with}\:\left(\mathrm{6}\right) \\ $$$$\mathrm{4}^{{th}} \:{answer}\:{x}=−\mathrm{1}−\sqrt{\frac{\mathrm{11}}{\mathrm{3}}},{y}=−\mathrm{1}+\sqrt{\frac{\mathrm{11}}{\mathrm{3}}} \\ $$
Commented by salaw2000 last updated on 19/Apr/19
thank you very much
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\: \\ $$
Answered by MJS last updated on 19/Apr/19
(1) y=((4x)/(3x−4)) (2) x^3 +y^3 −9xy=0 (1) in (2) ⇒ x^4 −4x^3 −((20)/3)x^2 +32x−((64)/3)=0 trying factors of ±((64)/3) we find x_1 =2; x_2 =4 ⇒ (x−2)(x−4)(x^2 +2x−(8/3))=0 ⇒ x_3 =−1−((√(33))/3); x_4 =−1+((√(33))/3) ⇒ ((x),(y) )∈{ ((2),(4) ), ((4),(2) ), (((−1−((√(33))/3))),((−1+((√(33))/3))) ) , (((−1−((√(33))/3))),((−1+((√(33))/3))) ) }
$$\left(\mathrm{1}\right)\:{y}=\frac{\mathrm{4}{x}}{\mathrm{3}{x}−\mathrm{4}} \\ $$$$\left(\mathrm{2}\right)\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} −\mathrm{9}{xy}=\mathrm{0} \\ $$$$\left(\mathrm{1}\right)\:\mathrm{in}\:\left(\mathrm{2}\right)\:\Rightarrow \\ $$$${x}^{\mathrm{4}} −\mathrm{4}{x}^{\mathrm{3}} −\frac{\mathrm{20}}{\mathrm{3}}{x}^{\mathrm{2}} +\mathrm{32}{x}−\frac{\mathrm{64}}{\mathrm{3}}=\mathrm{0} \\ $$$$\mathrm{trying}\:\mathrm{factors}\:\mathrm{of}\:\pm\frac{\mathrm{64}}{\mathrm{3}}\:\mathrm{we}\:\mathrm{find} \\ $$$${x}_{\mathrm{1}} =\mathrm{2};\:{x}_{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow \\ $$$$\left({x}−\mathrm{2}\right)\left({x}−\mathrm{4}\right)\left({x}^{\mathrm{2}} +\mathrm{2}{x}−\frac{\mathrm{8}}{\mathrm{3}}\right)=\mathrm{0} \\ $$$$\Rightarrow\:{x}_{\mathrm{3}} =−\mathrm{1}−\frac{\sqrt{\mathrm{33}}}{\mathrm{3}};\:{x}_{\mathrm{4}} =−\mathrm{1}+\frac{\sqrt{\mathrm{33}}}{\mathrm{3}} \\ $$$$\Rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\in\left\{\begin{pmatrix}{\mathrm{2}}\\{\mathrm{4}}\end{pmatrix},\:\begin{pmatrix}{\mathrm{4}}\\{\mathrm{2}}\end{pmatrix},\:\begin{pmatrix}{−\mathrm{1}−\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}}\\{−\mathrm{1}+\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}}\end{pmatrix}\:,\:\begin{pmatrix}{−\mathrm{1}−\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}}\\{−\mathrm{1}+\frac{\sqrt{\mathrm{33}}}{\mathrm{3}}}\end{pmatrix}\:\right\} \\ $$
Commented by Kunal12588 last updated on 19/Apr/19
Sir you are making me think why I did that ����
Commented by MJS last updated on 19/Apr/19
different paths lead to the same solutions − that′s a good thing!
$$\mathrm{different}\:\mathrm{paths}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{the}\:\mathrm{same}\:\mathrm{solutions}\:− \\ $$$$\mathrm{that}'\mathrm{s}\:\mathrm{a}\:\mathrm{good}\:\mathrm{thing}! \\ $$
Commented by salaw2000 last updated on 19/Apr/19
a very big thanks to you
$$\mathrm{a}\:\mathrm{very}\:\mathrm{big}\:\mathrm{thanks}\:\mathrm{to}\:\mathrm{you} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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