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Question Number 91977 by Power last updated on 04/May/20
Commented by jagoll last updated on 04/May/20
1−((5y)/x) = ((16)/x^2 ) , (y/x) = u , (1/x^2 ) = v ((y/x))^2 +((2y)/x) = (3/x^2 ) ⇒u^2 + 2u = 3v 16v+5u = 1⇒ v = ((1−5u)/(16)) ⇒u^2 +2u = ((3−15u)/(16)) 16u^2 +32u+15u−3=0 16u^2 +47u−3 = 0 (u+3)(16u−1)=0 { ((u=−3)),((u=(1/(16)) )) :}
$$\mathrm{1}−\frac{\mathrm{5y}}{\mathrm{x}}\:=\:\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }\:,\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{u}\:,\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{v} \\ $$$$\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2}} +\frac{\mathrm{2y}}{\mathrm{x}}\:=\:\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{u}^{\mathrm{2}} \:+\:\mathrm{2u}\:=\:\mathrm{3v} \\ $$$$\mathrm{16v}+\mathrm{5u}\:=\:\mathrm{1}\Rightarrow\:\mathrm{v}\:=\:\frac{\mathrm{1}−\mathrm{5u}}{\mathrm{16}} \\ $$$$\Rightarrow\mathrm{u}^{\mathrm{2}} +\mathrm{2u}\:=\:\frac{\mathrm{3}−\mathrm{15u}}{\mathrm{16}} \\ $$$$\mathrm{16u}^{\mathrm{2}} +\mathrm{32u}+\mathrm{15u}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{16u}^{\mathrm{2}} +\mathrm{47u}−\mathrm{3}\:=\:\mathrm{0} \\ $$$$\left(\mathrm{u}+\mathrm{3}\right)\left(\mathrm{16u}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\begin{cases}{\mathrm{u}=−\mathrm{3}}\\{\mathrm{u}=\frac{\mathrm{1}}{\mathrm{16}}\:}\end{cases} \\ $$$$ \\ $$
Commented by Rio Michael last updated on 04/May/20
second line seems correct to me 16 × 3 = 48
$$\mathrm{second}\:\mathrm{line}\:\mathrm{seems}\:\mathrm{correct}\:\mathrm{to}\:\mathrm{me} \\ $$$$\mathrm{16}\:×\:\mathrm{3}\:=\:\mathrm{48} \\ $$
Commented by Power last updated on 04/May/20
45xy no correct −15xy−32xy=−47xyu
$$\mathrm{45xy}\:\mathrm{no}\:\mathrm{correct}\:\:\:−\mathrm{15xy}−\mathrm{32xy}=−\mathrm{47xyu} \\ $$
Commented by jagoll last updated on 04/May/20
ig correct. huuray
$$\mathrm{ig}\:\mathrm{correct}.\:\mathrm{huuray} \\ $$
Commented by jagoll last updated on 04/May/20
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Commented by Power last updated on 04/May/20
???
$$??? \\ $$
Commented by jagoll last updated on 04/May/20
why??
$$\mathrm{why}?? \\ $$
Commented by jagoll last updated on 04/May/20
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Commented by john santu last updated on 04/May/20
yes. sir jagoll is correct
$$\mathrm{yes}.\:\mathrm{sir}\:\mathrm{jagoll}\:\mathrm{is}\:\mathrm{correct} \\ $$
Commented by Prithwish Sen 1 last updated on 04/May/20
let y=vx then (i/(ii)) we get ((1−5v)/(v^2 +2v)) = ((16)/3) ⇒ 16v^2 +47v −3=0 ⇒v =((−47±(√(2209+192)))/(32)) = ((−47±49)/(32)) = (1/(16)) , −3 ∴ y=k x=16k or y=−3k x=k or y=3k x=−k
$$\mathrm{let}\:\mathrm{y}=\mathrm{vx} \\ $$$$\mathrm{then}\:\:\frac{\mathrm{i}}{\mathrm{ii}}\:\mathrm{we}\:\mathrm{get} \\ $$$$\frac{\mathrm{1}−\mathrm{5v}}{\mathrm{v}^{\mathrm{2}} +\mathrm{2v}}\:=\:\frac{\mathrm{16}}{\mathrm{3}}\:\Rightarrow\:\mathrm{16v}^{\mathrm{2}} +\mathrm{47v}\:−\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\mathrm{v}\:=\frac{−\mathrm{47}\pm\sqrt{\mathrm{2209}+\mathrm{192}}}{\mathrm{32}}\:=\:\frac{−\mathrm{47}\pm\mathrm{49}}{\mathrm{32}}\:=\:\frac{\mathrm{1}}{\mathrm{16}}\:\:,\:\:−\mathrm{3} \\ $$$$\therefore\:\mathrm{y}=\mathrm{k}\:\:\mathrm{x}=\mathrm{16k} \\ $$$$\mathrm{or}\:\mathrm{y}=−\mathrm{3k}\:\:\mathrm{x}=\mathrm{k} \\ $$$$\mathrm{or}\:\mathrm{y}=\mathrm{3k}\:\:\mathrm{x}=−\mathrm{k} \\ $$
Answered by john santu last updated on 04/May/20
(x^2 /x^2 ) −((5y)/x) = ((16)/x^2 ) ⇒1−5((y/x))=((16)/x^2 ) say (y/x) = t & (1/x^2 ) = w 1−5t = 16w ⇒((y/x))^2 +((2y)/x) = (3/x^2 ) t^2 +2t = 3w t^2 +2t = 3(((1−5t)/(16))) 16t^2 +32t+15t−3=0 16t^2 +47t−3=0 (16t−1)(t+3)=0 case(1) t = −3 y=−3x ⇒9x^2 −6x^2 =3 x = ± 1 ∧y = ∓ 3 case(2) t=(1/(16))⇒x=16y y^2 +2(16y)y = 3 33y^2 = 3 ⇒y = ± (1/( (√(11)))) ∧x=±((16)/( (√(11))))
$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} }\:−\frac{\mathrm{5y}}{\mathrm{x}}\:=\:\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{1}−\mathrm{5}\left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\frac{\mathrm{16}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{say}\:\frac{\mathrm{y}}{\mathrm{x}}\:=\:\mathrm{t}\:\&\:\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\:=\:\mathrm{w} \\ $$$$\mathrm{1}−\mathrm{5t}\:=\:\mathrm{16w} \\ $$$$\Rightarrow\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^{\mathrm{2}} +\frac{\mathrm{2y}}{\mathrm{x}}\:=\:\frac{\mathrm{3}}{\mathrm{x}^{\mathrm{2}} } \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{2t}\:=\:\mathrm{3w} \\ $$$$\mathrm{t}^{\mathrm{2}} +\mathrm{2t}\:=\:\mathrm{3}\left(\frac{\mathrm{1}−\mathrm{5t}}{\mathrm{16}}\right) \\ $$$$\mathrm{16t}^{\mathrm{2}} +\mathrm{32t}+\mathrm{15t}−\mathrm{3}=\mathrm{0} \\ $$$$\mathrm{16t}^{\mathrm{2}} +\mathrm{47t}−\mathrm{3}=\mathrm{0} \\ $$$$\left(\mathrm{16t}−\mathrm{1}\right)\left(\mathrm{t}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{case}\left(\mathrm{1}\right)\:\mathrm{t}\:=\:−\mathrm{3} \\ $$$$\mathrm{y}=−\mathrm{3x}\:\Rightarrow\mathrm{9x}^{\mathrm{2}} −\mathrm{6x}^{\mathrm{2}} =\mathrm{3} \\ $$$$\mathrm{x}\:=\:\pm\:\mathrm{1}\:\wedge\mathrm{y}\:=\:\mp\:\mathrm{3} \\ $$$$\mathrm{case}\left(\mathrm{2}\right)\:\mathrm{t}=\frac{\mathrm{1}}{\mathrm{16}}\Rightarrow\mathrm{x}=\mathrm{16y} \\ $$$$\mathrm{y}^{\mathrm{2}} +\mathrm{2}\left(\mathrm{16y}\right)\mathrm{y}\:=\:\mathrm{3} \\ $$$$\mathrm{33y}^{\mathrm{2}} \:=\:\mathrm{3}\:\Rightarrow\mathrm{y}\:=\:\pm\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}\:\wedge\mathrm{x}=\pm\frac{\mathrm{16}}{\:\sqrt{\mathrm{11}}} \\ $$$$ \\ $$
Commented by jagoll last updated on 04/May/20
i agree sir
$$\mathrm{i}\:\mathrm{agree}\:\mathrm{sir} \\ $$
Answered by $@ty@m123 last updated on 04/May/20
((x^2 −5xy)/(y^2 +2xy)) = ((16)/3) (((x^2 /y^2 )−5((x/y)))/(1+2((x/y))))=((16)/3) ((t^2 −5t)/(1+2t))=((16)/3) { where t=(x/y) 3(t^2 −5t)=16(1+2t) 3t^2 −47t−16=0 t=((47±(√(2209+192)))/6) t=((47±49)/6) t=((−2)/6), ((96)/6) t=((−1)/3), 16 ⇒x=((−y)/3), x=16y Case 1. x=((−y)/3) (((−y)/3))^2 −5(((−y)/3))y=16 y^2 +15y^2 =144 16y^2 =144 y=±3, x=∓1 Case 2. x=16y (16y)^2 −5(16y)y=16 16y^2 −5y^2 =1 11y^2 =1 y=±(1/( (√(11)))),x=±((16)/( (√(11))))
$$\frac{{x}^{\mathrm{2}} −\mathrm{5}{xy}}{{y}^{\mathrm{2}} +\mathrm{2}{xy}}\:=\:\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\frac{\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }−\mathrm{5}\left(\frac{{x}}{{y}}\right)}{\mathrm{1}+\mathrm{2}\left(\frac{{x}}{{y}}\right)}=\frac{\mathrm{16}}{\mathrm{3}} \\ $$$$\frac{{t}^{\mathrm{2}} −\mathrm{5}{t}}{\mathrm{1}+\mathrm{2}{t}}=\frac{\mathrm{16}}{\mathrm{3}}\:\:\left\{\:{where}\:{t}=\frac{{x}}{{y}}\right. \\ $$$$\mathrm{3}\left({t}^{\mathrm{2}} −\mathrm{5}{t}\right)=\mathrm{16}\left(\mathrm{1}+\mathrm{2}{t}\right) \\ $$$$\mathrm{3}{t}^{\mathrm{2}} −\mathrm{47}{t}−\mathrm{16}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{47}\pm\sqrt{\mathrm{2209}+\mathrm{192}}}{\mathrm{6}} \\ $$$${t}=\frac{\mathrm{47}\pm\mathrm{49}}{\mathrm{6}} \\ $$$${t}=\frac{−\mathrm{2}}{\mathrm{6}},\:\frac{\mathrm{96}}{\mathrm{6}} \\ $$$${t}=\frac{−\mathrm{1}}{\mathrm{3}},\:\mathrm{16} \\ $$$$\Rightarrow{x}=\frac{−{y}}{\mathrm{3}},\:{x}=\mathrm{16}{y} \\ $$$${Case}\:\mathrm{1}.\:{x}=\frac{−{y}}{\mathrm{3}} \\ $$$$\left(\frac{−{y}}{\mathrm{3}}\right)^{\mathrm{2}} −\mathrm{5}\left(\frac{−{y}}{\mathrm{3}}\right){y}=\mathrm{16} \\ $$$${y}^{\mathrm{2}} +\mathrm{15}{y}^{\mathrm{2}} =\mathrm{144} \\ $$$$\mathrm{16}{y}^{\mathrm{2}} =\mathrm{144} \\ $$$${y}=\pm\mathrm{3},\:{x}=\mp\mathrm{1} \\ $$$${Case}\:\mathrm{2}.\:{x}=\mathrm{16}{y} \\ $$$$\left(\mathrm{16}{y}\right)^{\mathrm{2}} −\mathrm{5}\left(\mathrm{16}{y}\right){y}=\mathrm{16} \\ $$$$\mathrm{16}{y}^{\mathrm{2}} −\mathrm{5}{y}^{\mathrm{2}} =\mathrm{1} \\ $$$$\mathrm{11}{y}^{\mathrm{2}} =\mathrm{1} \\ $$$${y}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}},{x}=\pm\frac{\mathrm{16}}{\:\sqrt{\mathrm{11}}} \\ $$
Commented by jagoll last updated on 04/May/20
yes
$$\mathrm{yes} \\ $$
Answered by Rasheed.Sindhi last updated on 04/May/20
•(i)×3: 3x^2 −15xy=48 •(ii)×16: 16y^2 +32xy=48 (i)×3 − (ii)×16: 3x^2 −47xy−16y^2 =0 (x−16y)(3x+y)=0 x−16y=0 ∣ 3x+y=0 x=16y ∣ y=−3x ^• When x=16y y^2 +2xy=3⇒y^2 +32y^2 =3 y=±(1/( (√(11))))=±((√(11))/(11)) x=±((16(√(11)))/(11)) (x,y)=(±((16(√(11)))/(11)) , ±((√(11))/(11))) ^• When y=−3x x^2 −5xy=16⇒x^2 +15x^2 =16 x=±1 y=∓3 (x,y)=(±1 , ∓3) (x,y)=(±((16(√(11)))/(11)) , ±((√(11))/(11))) ∣ (±1, ∓3)
$$\bullet\left({i}\right)×\mathrm{3}:\:\:\:\:\:\:\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{15}{xy}=\mathrm{48}\:\: \\ $$$$\bullet\left({ii}\right)×\mathrm{16}:\:\:\mathrm{16}{y}^{\mathrm{2}} +\mathrm{32}{xy}=\mathrm{48} \\ $$$$\:\:\:\:\left({i}\right)×\mathrm{3}\:\:−\:\:\left({ii}\right)×\mathrm{16}: \\ $$$$\:\:\:\:\mathrm{3}{x}^{\mathrm{2}} −\mathrm{47}{xy}−\mathrm{16}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\:\:\left({x}−\mathrm{16}{y}\right)\left(\mathrm{3}{x}+{y}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:{x}−\mathrm{16}{y}=\mathrm{0}\:\:\mid\:\:\:\mathrm{3}{x}+{y}=\mathrm{0} \\ $$$$\:\:\:\:\:{x}=\mathrm{16}{y}\:\:\:\:\mid\:\:\:\:\:{y}=−\mathrm{3}{x} \\ $$$$\:^{\bullet} {When}\:{x}=\mathrm{16}{y} \\ $$$$\:\:\:\:\:\:{y}^{\mathrm{2}} +\mathrm{2}{xy}=\mathrm{3}\Rightarrow{y}^{\mathrm{2}} +\mathrm{32}{y}^{\mathrm{2}} =\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}=\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\pm\frac{\mathrm{16}\sqrt{\mathrm{11}}}{\mathrm{11}} \\ $$$$\:\:\:\:\:\left({x},{y}\right)=\left(\pm\frac{\mathrm{16}\sqrt{\mathrm{11}}}{\mathrm{11}}\:,\:\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{11}}\right) \\ $$$$\:^{\bullet} {When}\:{y}=−\mathrm{3}{x} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{2}} −\mathrm{5}{xy}=\mathrm{16}\Rightarrow{x}^{\mathrm{2}} +\mathrm{15}{x}^{\mathrm{2}} =\mathrm{16} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}=\pm\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}=\mp\mathrm{3} \\ $$$$\:\:\:\:\:\:\left({x},{y}\right)=\left(\pm\mathrm{1}\:,\:\mp\mathrm{3}\right) \\ $$$$ \\ $$$$\left({x},{y}\right)=\left(\pm\frac{\mathrm{16}\sqrt{\mathrm{11}}}{\mathrm{11}}\:,\:\pm\frac{\sqrt{\mathrm{11}}}{\mathrm{11}}\right)\:\:\mid\:\:\left(\pm\mathrm{1},\:\mp\mathrm{3}\right) \\ $$
Commented by Power last updated on 04/May/20
thanks
$$\mathrm{thanks} \\ $$
Commented by Rasheed.Sindhi last updated on 04/May/20
Thsnks sir! I′ve corrected.
$${Thsnks}\:{sir}!\:{I}'{ve}\:{corrected}. \\ $$
Commented by Rasheed.Sindhi last updated on 04/May/20
fine sir! Happy to know that you′re still remembering me!
$${fine}\:{sir}! \\ $$$${Happy}\:{to}\:{know}\:{that}\:{you}'{re}\:{still} \\ $$$${remembering}\:{me}! \\ $$
Commented by Prithwish Sen 1 last updated on 04/May/20
Rasheed sir how are you ?
$$\mathrm{Rasheed}\:\mathrm{sir}\:\mathrm{how}\:\mathrm{are}\:\mathrm{you}\:? \\ $$
Commented by jagoll last updated on 04/May/20
wrong in line 7
$$\mathrm{wrong}\:\mathrm{in}\:\mathrm{line}\:\mathrm{7} \\ $$
Commented by Prithwish Sen 1 last updated on 04/May/20
why not sir ? we had a good time in this forum.
$$\mathrm{why}\:\mathrm{not}\:\mathrm{sir}\:?\:\mathrm{we}\:\mathrm{had}\:\mathrm{a}\:\mathrm{good}\:\mathrm{time}\:\:\mathrm{in}\:\mathrm{this}\:\mathrm{forum}. \\ $$
Commented by $@ty@m123 last updated on 04/May/20
cutest of all solutions.
$${cutest}\:{of}\:{all}\:{solutions}. \\ $$
Commented by Rasheed.Sindhi last updated on 04/May/20
ThanX sir!
$${Than}\mathcal{X}\:{sir}! \\ $$
Commented by Rasheed.Sindhi last updated on 04/May/20
To Prithwish Sen, Of course sir!
$$\mathcal{T}{o}\:{Prithwish}\:{Sen}, \\ $$$$\mathcal{O}{f}\:{course}\:{sir}! \\ $$
Answered by MJS last updated on 04/May/20
{ ((y=((x^2 −16)/(5x)))),((⇒ x^4 −((267)/(11))x^2 +((256)/(11))=0)) :} ⇒ ((x),(y) ) = (((±1)),((∓3)) ) ∨ ((x),(y) ) = (((±((16)/( (√(11)))))),((±(1/( (√(11)))))) )
$$\begin{cases}{{y}=\frac{{x}^{\mathrm{2}} −\mathrm{16}}{\mathrm{5}{x}}}\\{\Rightarrow\:{x}^{\mathrm{4}} −\frac{\mathrm{267}}{\mathrm{11}}{x}^{\mathrm{2}} +\frac{\mathrm{256}}{\mathrm{11}}=\mathrm{0}}\end{cases} \\ $$$$\Rightarrow\:\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\pm\mathrm{1}}\\{\mp\mathrm{3}}\end{pmatrix}\:\vee\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}\:=\begin{pmatrix}{\pm\frac{\mathrm{16}}{\:\sqrt{\mathrm{11}}}}\\{\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{11}}}}\end{pmatrix} \\ $$

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