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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

������

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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