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Question Number 109738 by Khalmohmmad last updated on 25/Aug/20

(((x/y)−(y/x))/((x^2 +2xy+y^2 )/(x^2 −y^2 )))=(9/4) (x/y)=?

$$\frac{\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}=? \\ $$

Commented by nimnim last updated on 25/Aug/20

Sol: (((x/y)−(y/x))/((x^2 +2xy+y^2 )/(x^2 −y^2 )))=(9/4) ⇒ ((x^2 −y^2 )/(xy))×((x^2 −y^2 )/((x+y)^2 ))=(9/4) ⇒(((x−y)^2 )/(xy))=(9/4) ⇒(((x−y)^2 )/(4xy))=(9/(16)).........(i) ⇒(((x−y)^2 +4xy)/(4xy))=((9+16)/(16)) {componendo} ⇒(((x+y)^2 )/(4xy))=((25)/(16))........(ii) (ii)÷(i)⇒(((x+y)^2 )/((x−y)^2 ))=((25)/9) ⇒((x+y)/(x−y))=(5/3) or ((x+y)/(x−y))=−(5/3) by componendo and dividendo ⇒((2x)/(2y))=(8/2) or ((2x)/(2y))=((−2)/(−8)) ⇒(x/y)=4 or (x/y)=(1/4)

$$\mathrm{Sol}:\:\frac{\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\:\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\mathrm{xy}}×\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} }{\mathrm{xy}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} }{\mathrm{4xy}}=\frac{\mathrm{9}}{\mathrm{16}}.........\left(\mathrm{i}\right) \\ $$$$\Rightarrow\frac{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} +\mathrm{4xy}}{\mathrm{4xy}}=\frac{\mathrm{9}+\mathrm{16}}{\mathrm{16}}\:\:\left\{\mathrm{componendo}\right\} \\ $$$$\Rightarrow\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }{\mathrm{4xy}}=\frac{\mathrm{25}}{\mathrm{16}}........\left(\mathrm{ii}\right) \\ $$$$\left(\mathrm{ii}\right)\boldsymbol{\div}\left(\mathrm{i}\right)\Rightarrow\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{x}−\mathrm{y}\right)^{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{9}} \\ $$$$\Rightarrow\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\mathrm{or}\:\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}=−\frac{\mathrm{5}}{\mathrm{3}} \\ $$$$\mathrm{by}\:\mathrm{componendo}\:\mathrm{and}\:\mathrm{dividendo} \\ $$$$\Rightarrow\frac{\mathrm{2x}}{\mathrm{2y}}=\frac{\mathrm{8}}{\mathrm{2}}\:\:\:\:\:\mathrm{or}\:\:\:\frac{\mathrm{2x}}{\mathrm{2y}}=\frac{−\mathrm{2}}{−\mathrm{8}} \\ $$$$\Rightarrow\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{4}\:\:\:\mathrm{or}\:\:\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$

Answered by Rasheed.Sindhi last updated on 25/Aug/20

(((x/y)−(y/x))/((x^2 +2xy+y^2 )/(x^2 −y^2 )))=(9/4) (( (x/y)−(y/x) )/((x+y)/(x−y)))=(9/4) (( (x/y)−(y/x) )/(( ((x+y)/y) )/((x−y)/y)))=(9/4) (( (x/y)−(y/x) )/(( (x/y)+1 )/((x/y)−1)))=(9/4) Let (x/y)=u⇒(y/x)=(1/u) (( u−(1/u) )/((u+1)/(u−1)))=(9/4) 4(((u^2 −1)/u))=9(((u+1)/(u−1))) 4(u^2 −1)(u−1)=9u(u+1) 4(u−1)^2 =9u 4u^2 −17u+4=0 4u^2 −16u−u+4 4u(u−4)−1(u−4) (4u−1)(u−4)=0 u=1/4 ∣ u=4 (x/y)=1/4 ∣ (x/y)=4

$$\frac{\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{\:\:\:\:\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}\:\:\:}{\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}\:\:\:\:\:}{\frac{\:\:\:\:\:\frac{\mathrm{x}+\mathrm{y}}{\mathrm{y}}\:\:\:}{\frac{\mathrm{x}−\mathrm{y}}{\mathrm{y}}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{\:\:\:\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}\:\:\:\:\:}{\frac{\:\:\:\:\:\frac{\mathrm{x}}{\mathrm{y}}+\mathrm{1}\:\:\:}{\frac{\mathrm{x}}{\mathrm{y}}−\mathrm{1}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$${Let}\:\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{u}\Rightarrow\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{1}}{\mathrm{u}} \\ $$$$\frac{\:\:\:\:\:\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\:\:\:\:\:}{\frac{\mathrm{u}+\mathrm{1}}{\mathrm{u}−\mathrm{1}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{u}}\right)=\mathrm{9}\left(\frac{\mathrm{u}+\mathrm{1}}{\mathrm{u}−\mathrm{1}}\right) \\ $$$$\mathrm{4}\left(\mathrm{u}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{u}−\mathrm{1}\right)=\mathrm{9u}\left(\mathrm{u}+\mathrm{1}\right) \\ $$$$\mathrm{4}\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9u} \\ $$$$\mathrm{4u}^{\mathrm{2}} −\mathrm{17u}+\mathrm{4}=\mathrm{0} \\ $$$$\mathrm{4u}^{\mathrm{2}} −\mathrm{16u}−\mathrm{u}+\mathrm{4} \\ $$$$\mathrm{4u}\left(\mathrm{u}−\mathrm{4}\right)−\mathrm{1}\left(\mathrm{u}−\mathrm{4}\right) \\ $$$$\left(\mathrm{4u}−\mathrm{1}\right)\left(\mathrm{u}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{u}=\mathrm{1}/\mathrm{4}\:\mid\:\mathrm{u}=\mathrm{4} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{1}/\mathrm{4}\:\mid\:\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{4}\:\:\:\:\:\:\:\:\:\:\:\: \\ $$

Commented by bobhans last updated on 25/Aug/20

4u^2 −17u+4 = (u−((16)/4))(u−(1/4)) =(u−4)(4u−1) sir

$$\mathrm{4}{u}^{\mathrm{2}} −\mathrm{17}{u}+\mathrm{4}\:=\:\left({u}−\frac{\mathrm{16}}{\mathrm{4}}\right)\left({u}−\frac{\mathrm{1}}{\mathrm{4}}\right) \\ $$$$=\left({u}−\mathrm{4}\right)\left(\mathrm{4}{u}−\mathrm{1}\right)\:{sir} \\ $$

Commented by Rasheed.Sindhi last updated on 25/Aug/20

Thanks Bob for mentioning mistake!

$$\mathcal{T}{hanks}\:{Bob}\:{for}\:{mentioning} \\ $$$${mistake}! \\ $$

Answered by Rasheed.Sindhi last updated on 25/Aug/20

(( (x/y)−(y/x) )/(( x^2 +2xy+y^2 )/(x^2 −y^2 )))=(9/4) (x/y)=u⇒x=uy (( ((uy)/y)−(y/(uy)) )/(( (uy)^2 +2(uy)y+y^2 )/((uy)^2 −y^2 )))=(9/4) (( u−(1/u) )/(( u^2 y^2 +2uy^2 +y^2 )/(u^2 y^2 −y^2 )))=(9/4) (( u−(1/u) )/(( u^2 +2u+1)/(u^2 −1)))=(9/4) 4(((u^2 −1)/u))=9(((u+1)/(u−1))) 4(u−1)^2 (u+1)=9u(u+1) 4(u−1)^2 =9u 4u^2 −17u+4=0 (4u−1)(u−4)=0 u=1/4 ∣ u=4 (x/y)=1/4 ∣ (x/y)=4

$$\:\:\frac{\:\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}\:\:}{\frac{\:\:\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{u}\Rightarrow\mathrm{x}=\mathrm{uy} \\ $$$$\:\:\frac{\:\frac{\mathrm{uy}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{uy}}\:\:}{\frac{\:\:\left(\mathrm{uy}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{uy}\right)\mathrm{y}+\mathrm{y}^{\mathrm{2}} }{\left(\mathrm{uy}\right)^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:\frac{\:\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\:\:}{\frac{\:\:\mathrm{u}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} +\mathrm{2uy}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{u}^{\mathrm{2}} \mathrm{y}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\:\:\frac{\:\:\:\mathrm{u}−\frac{\mathrm{1}}{\mathrm{u}}\:\:\:}{\frac{\:\:\mathrm{u}^{\mathrm{2}} +\mathrm{2u}+\mathrm{1}}{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\mathrm{4}\left(\frac{\mathrm{u}^{\mathrm{2}} −\mathrm{1}}{\mathrm{u}}\right)=\mathrm{9}\left(\frac{\mathrm{u}+\mathrm{1}}{\mathrm{u}−\mathrm{1}}\right) \\ $$$$\mathrm{4}\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{u}+\mathrm{1}\right)=\mathrm{9u}\left(\mathrm{u}+\mathrm{1}\right) \\ $$$$\mathrm{4}\left(\mathrm{u}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9u} \\ $$$$\mathrm{4u}^{\mathrm{2}} −\mathrm{17u}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{4u}−\mathrm{1}\right)\left(\mathrm{u}−\mathrm{4}\right)=\mathrm{0} \\ $$$$\mathrm{u}=\mathrm{1}/\mathrm{4}\:\mid\:\mathrm{u}=\mathrm{4} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{1}/\mathrm{4}\:\mid\:\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{4} \\ $$

Answered by ajfour last updated on 25/Aug/20

t=(x/y) (((t−(1/t)))/([(((t+1)^2 )/(t^2 −1))]))=(9/4) ⇒ 4(t^2 −1)^2 =9t(t+1)^2 t≠−1 ⇒ 4(t−1)^2 =9t 4t^2 −17t+4=0 t=((17)/8)±((√(289−64))/8) t = ((17±15)/8) = 4, (1/4) ⇒ t=(x/y) = (1/4), 4 .

$${t}=\frac{{x}}{{y}} \\ $$$$\frac{\left({t}−\frac{\mathrm{1}}{{t}}\right)}{\left[\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{{t}^{\mathrm{2}} −\mathrm{1}}\right]}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\mathrm{4}\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9}{t}\left({t}+\mathrm{1}\right)^{\mathrm{2}} \:\:\: \\ $$$$\:{t}\neq−\mathrm{1} \\ $$$$\Rightarrow\:\:\mathrm{4}\left({t}−\mathrm{1}\right)^{\mathrm{2}} =\mathrm{9}{t} \\ $$$$\mathrm{4}{t}^{\mathrm{2}} −\mathrm{17}{t}+\mathrm{4}=\mathrm{0} \\ $$$${t}=\frac{\mathrm{17}}{\mathrm{8}}\pm\frac{\sqrt{\mathrm{289}−\mathrm{64}}}{\mathrm{8}} \\ $$$$\:{t}\:=\:\frac{\mathrm{17}\pm\mathrm{15}}{\mathrm{8}}\:=\:\mathrm{4},\:\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\Rightarrow\:\:\:{t}=\frac{{x}}{{y}}\:=\:\frac{\mathrm{1}}{\mathrm{4}},\:\mathrm{4}\:. \\ $$

Answered by 1549442205PVT last updated on 25/Aug/20

((x^2 +2xy+y^2 )/(x^2 −y^2 ))=(((x+y)^2 )/((x+y)(x−y)))=((x+y)/(x−y)),so (((x/y)−(y/x))/((x^2 +2xy+y^2 )/(x^2 −y^2 )))=(9/4)⇔(((x/y)−(y/x))/((x+y)/(x−y)))=(9/4) ⇔(x/y)−(y/x)=(9/4)×((x+y)/(x−y))⇔((x^2 −y^2 )/(xy))=(9/4).((x+y)/(x−y)) ⇔(((x+y)(x−y))/(xy))×((x−y)/(x+y))=(9/4) ⇔((x^2 −2xy+y^2 )/(xy))=(9/4)⇔(x/y)+(y/x)−2=(9/4) ⇔(x/y)+(y/x)=((17)/4).Put (x/y)=t we have t+(1/t)=((17)/4)⇔4t^2 −17t+4=0 Δ=17^2 −64=289−64=225=15^2 ⇔t=((17±15)/8)∈{4,(1/4)} Thu s, (x/y)∈{4;(1/4)}

$$ \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }=\frac{\left(\mathrm{x}+\mathrm{y}\right)^{\mathrm{2}} }{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}−\mathrm{y}\right)}=\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}},\mathrm{so} \\ $$$$\frac{\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }}=\frac{\mathrm{9}}{\mathrm{4}}\Leftrightarrow\frac{\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}}{\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{x}}{\mathrm{y}}−\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{9}}{\mathrm{4}}×\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}}\Leftrightarrow\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{y}^{\mathrm{2}} }{\mathrm{xy}}=\frac{\mathrm{9}}{\mathrm{4}}.\frac{\mathrm{x}+\mathrm{y}}{\mathrm{x}−\mathrm{y}} \\ $$$$\Leftrightarrow\frac{\left(\mathrm{x}+\mathrm{y}\right)\left(\mathrm{x}−\mathrm{y}\right)}{\mathrm{xy}}×\frac{\mathrm{x}−\mathrm{y}}{\mathrm{x}+\mathrm{y}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{xy}}=\frac{\mathrm{9}}{\mathrm{4}}\Leftrightarrow\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}−\mathrm{2}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\Leftrightarrow\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{17}}{\mathrm{4}}.\mathrm{Put}\:\frac{\mathrm{x}}{\mathrm{y}}=\mathrm{t}\:\mathrm{we}\:\mathrm{have} \\ $$$$\mathrm{t}+\frac{\mathrm{1}}{\mathrm{t}}=\frac{\mathrm{17}}{\mathrm{4}}\Leftrightarrow\mathrm{4t}^{\mathrm{2}} −\mathrm{17t}+\mathrm{4}=\mathrm{0} \\ $$$$\Delta=\mathrm{17}^{\mathrm{2}} −\mathrm{64}=\mathrm{289}−\mathrm{64}=\mathrm{225}=\mathrm{15}^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{t}=\frac{\mathrm{17}\pm\mathrm{15}}{\mathrm{8}}\in\left\{\mathrm{4},\frac{\mathrm{1}}{\mathrm{4}}\right\} \\ $$$$\:\boldsymbol{\mathrm{Thu}}\:\boldsymbol{\mathrm{s}},\:\:\frac{\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{y}}}\in\left\{\mathrm{4};\frac{\mathrm{1}}{\mathrm{4}}\right\} \\ $$

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