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Question Number 137982 by mey3nipaba last updated on 08/Apr/21
y^2 −x^2 =5(y−x)^2 . Find x:y
$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{5}\left({y}−{x}\right)^{\mathrm{2}} .\:{Find}\:{x}:{y} \\ $$
Commented by mey3nipaba last updated on 08/Apr/21
please help
$${please}\:{help} \\ $$
Answered by Dwaipayan Shikari last updated on 08/Apr/21
((y^2 −x^2 )/((y−x)^2 ))=5⇒(((y−x)(y+x))/((y−x)(y−x)))=5 ⇒((y+x)/(y−x))=5⇒((y+x+(y−x))/(y+x−(y−x)))=((5+1)/(5−1))⇒(y/x)=(6/4) ⇒(x/y)=(2/3)
$$\frac{{y}^{\mathrm{2}} −{x}^{\mathrm{2}} }{\left({y}−{x}\right)^{\mathrm{2}} }=\mathrm{5}\Rightarrow\frac{\left({y}−{x}\right)\left({y}+{x}\right)}{\left({y}−{x}\right)\left({y}−{x}\right)}=\mathrm{5} \\ $$$$\Rightarrow\frac{{y}+{x}}{{y}−{x}}=\mathrm{5}\Rightarrow\frac{{y}+{x}+\left({y}−{x}\right)}{{y}+{x}−\left({y}−{x}\right)}=\frac{\mathrm{5}+\mathrm{1}}{\mathrm{5}−\mathrm{1}}\Rightarrow\frac{{y}}{{x}}=\frac{\mathrm{6}}{\mathrm{4}}\:\: \\ $$$$\Rightarrow\frac{{x}}{{y}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$
Commented by mey3nipaba last updated on 08/Apr/21
thanks so much. i had the answer but wanted to confirm
$${thanks}\:{so}\:{much}.\:{i}\:{had}\:{the}\:{answer}\:{but}\:{wanted}\:{to}\:{confirm} \\ $$
Answered by Rasheed.Sindhi last updated on 08/Apr/21
An Alternate Way (x/y)=k⇒x=yk y^2 −x^2 =5(y−x)^2 ⇒y^2 −(yk)^2 =5(y−yk)^2 y^2 (1−k^2 )=5y^2 (1−k)^2 y^2 (1−k)(1+k)−5y^2 (1−k)^2 =0 (1−k)( y^2 (1+k)−5y^2 (1−k) )=0 k=1 ∣ y^2 (1+k)=5y^2 (1−k) x:y=1:1 ∣ ((1+k)/(1−k))=((5y^2 )/y^2 )=5 ∣ 1+k=5−5k ∣ 6k=4⇒k=2/3 ∣ x:y=2:3
$${An}\:{Alternate}\:{Way} \\ $$$$\frac{{x}}{{y}}={k}\Rightarrow{x}={yk} \\ $$$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{5}\left({y}−{x}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{y}^{\mathrm{2}} −\left({yk}\right)^{\mathrm{2}} =\mathrm{5}\left({y}−{yk}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{y}^{\mathrm{2}} \left(\mathrm{1}−{k}^{\mathrm{2}} \right)=\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{1}−{k}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:{y}^{\mathrm{2}} \left(\mathrm{1}−{k}\right)\left(\mathrm{1}+{k}\right)−\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{1}−{k}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\:\:\:\left(\mathrm{1}−{k}\right)\left(\:{y}^{\mathrm{2}} \left(\mathrm{1}+{k}\right)−\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{1}−{k}\right)\:\right)=\mathrm{0} \\ $$$$\:\:\:\:{k}=\mathrm{1}\:\mid\:\:{y}^{\mathrm{2}} \left(\mathrm{1}+{k}\right)=\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{1}−{k}\right)\: \\ $$$$\:\:{x}:{y}=\mathrm{1}:\mathrm{1}\:\:\mid\:\frac{\mathrm{1}+{k}}{\mathrm{1}−{k}}=\frac{\mathrm{5}{y}^{\mathrm{2}} }{{y}^{\mathrm{2}} }=\mathrm{5} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\mathrm{1}+{k}=\mathrm{5}−\mathrm{5}{k} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:\mathrm{6}{k}=\mathrm{4}\Rightarrow{k}=\mathrm{2}/\mathrm{3} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\:\:\:\:{x}:{y}=\mathrm{2}:\mathrm{3} \\ $$
Answered by nadovic last updated on 08/Apr/21
y^2 − x^2 = 5y^2 − 10xy + 5x^2 6x^2 − 10xy + 4y^2 = 0 2(3x^2 − 5xy + 2y^2 ) = 0 2(x − y)(3x − 2y) = 0 (x/y) = 1 or (x/y) = (2/3) x:y = 1:1 or x:y = 2:3
$${y}^{\mathrm{2}} \:−\:{x}^{\mathrm{2}} \:=\:\mathrm{5}{y}^{\mathrm{2}} \:−\:\mathrm{10}{xy}\:+\:\mathrm{5}{x}^{\mathrm{2}} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} \:−\:\mathrm{10}{xy}\:+\:\mathrm{4}{y}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\mathrm{2}\left(\mathrm{3}{x}^{\mathrm{2}} \:−\:\mathrm{5}{xy}\:+\:\mathrm{2}{y}^{\mathrm{2}} \right)\:=\:\mathrm{0} \\ $$$$\mathrm{2}\left({x}\:−\:{y}\right)\left(\mathrm{3}{x}\:−\:\mathrm{2}{y}\right)\:=\:\mathrm{0} \\ $$$$\frac{{x}}{{y}}\:=\:\mathrm{1}\:\:{or}\:\:\frac{{x}}{{y}}\:=\:\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}:{y}\:=\:\mathrm{1}:\mathrm{1}\:\:\:{or}\:\:\:{x}:{y}\:=\:\mathrm{2}:\mathrm{3} \\ $$
Answered by Rasheed.Sindhi last updated on 09/Apr/21
y^2 −x^2 =5(y−x)^2 . Find x:y y^2 (1−(x^2 /y^2 ))=5y^2 (1−(x/y))^2 y≠0⇒(1−(x^2 /y^2 ))=5(1−(x/y))^2 (x/y)=u⇒(1−u^2 )=5(1−u)^2 (1−u)(1+u)−5(1−u)^2 =0 (1−u)(1+u−5+5u)=0 (1−u)(6u−4)=0 u=1 ∣ u=(2/3) x:y=1:1 ∣ x:y=2:3
$${y}^{\mathrm{2}} −{x}^{\mathrm{2}} =\mathrm{5}\left({y}−{x}\right)^{\mathrm{2}} .\:{Find}\:{x}:{y} \\ $$$${y}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\right)=\mathrm{5}{y}^{\mathrm{2}} \left(\mathrm{1}−\frac{{x}}{{y}}\right)^{\mathrm{2}} \\ $$$${y}\neq\mathrm{0}\Rightarrow\left(\mathrm{1}−\frac{{x}^{\mathrm{2}} }{{y}^{\mathrm{2}} }\right)=\mathrm{5}\left(\mathrm{1}−\frac{{x}}{{y}}\right)^{\mathrm{2}} \:\: \\ $$$$\frac{{x}}{{y}}={u}\Rightarrow\left(\mathrm{1}−{u}^{\mathrm{2}} \right)=\mathrm{5}\left(\mathrm{1}−{u}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}\right)−\mathrm{5}\left(\mathrm{1}−{u}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{1}−{u}\right)\left(\mathrm{1}+{u}−\mathrm{5}+\mathrm{5}{u}\right)=\mathrm{0} \\ $$$$\left(\mathrm{1}−{u}\right)\left(\mathrm{6}{u}−\mathrm{4}\right)=\mathrm{0} \\ $$$${u}=\mathrm{1}\:\mid\:\:{u}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$$${x}:{y}=\mathrm{1}:\mathrm{1}\:\:\mid\:\:{x}:{y}=\mathrm{2}:\mathrm{3} \\ $$
Answered by MJS_new last updated on 09/Apr/21
(y−x)(y+x)=5(y−x)^2 (1) y−x=0 ⇒ (x/y)=1 (2) y+x=5(y−x) ⇒ (x/y)=(2/3)
$$\left({y}−{x}\right)\left({y}+{x}\right)=\mathrm{5}\left({y}−{x}\right)^{\mathrm{2}} \\ $$$$\left(\mathrm{1}\right)\:{y}−{x}=\mathrm{0}\:\:\Rightarrow\:\frac{{x}}{{y}}=\mathrm{1} \\ $$$$\left(\mathrm{2}\right)\:{y}+{x}=\mathrm{5}\left({y}−{x}\right)\:\Rightarrow\:\frac{{x}}{{y}}=\frac{\mathrm{2}}{\mathrm{3}} \\ $$

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