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Question Number 137982 by mey3nipaba last updated on 08/Apr/21

$$y^2-x^2=5{(y-x)}^2.Findx:y$$

Commented by mey3nipaba last updated on 08/Apr/21

$$pleasehelp$$

Answered by Dwaipayan Shikari last updated on 08/Apr/21

$$\frac{y^2-x^2}{{(y-x)}^2}=5\Rightarrow \frac{(y-x)(y+x)}{(y-x)(y-x)}=5$$$$\Rightarrow \frac{y+x}{y-x}=5\Rightarrow \frac{y+x+(y-x)}{y+x-(y-x)}=\frac{5+1}{5-1}\Rightarrow \frac{y}{x}=\frac{6}{4}$$$$\Rightarrow \frac{x}{y}=\frac{2}{3}$$

Commented by mey3nipaba last updated on 08/Apr/21

$$thankssomuch.ihadtheanswerbutwantedtoconfirm$$

Answered by Rasheed.Sindhi last updated on 08/Apr/21

$$AnAlternateWay$$$$\frac{x}{y}=k\Rightarrow x=yk$$$$y^2-x^2=5{(y-x)}^2$$$$\Rightarrow y^2-{\left(yk\right)}^2=5{(y-yk)}^2$$$$y^2(1-k^2)=5y^2{(1-k)}^2$$$$y^2(1-k)(1+k)-5y^2{(1-k)}^2=0$$$$(1-k)(y^2(1+k)-5y^2(1-k))=0$$$$k=1\mid y^2(1+k)=5y^2(1-k)$$$$x:y=1:1\mid \frac{1+k}{1-k}=\frac{5y^2}{y^2}=5$$$$\mid 1+k=5-5k$$$$\mid 6k=4\Rightarrow k=2/3$$$$\mid x:y=2:3$$

Answered by nadovic last updated on 08/Apr/21

$$y^2-x^2=5y^2-10xy+5x^2$$$$6x^2-10xy+4y^2=0$$$$2(3x^2-5xy+2y^2)=0$$$$2(x-y)(3x-2y)=0$$$$\frac{x}{y}=1or\frac{x}{y}=\frac{2}{3}$$$$x:y=1:1orx:y=2:3$$

Answered by Rasheed.Sindhi last updated on 09/Apr/21

$$y^2-x^2=5{(y-x)}^2.Findx:y$$$$y^2(1-\frac{x^2}{y^2})=5y^2{(1-\frac{x}{y})}^2$$$$y\ne 0\Rightarrow (1-\frac{x^2}{y^2})=5{(1-\frac{x}{y})}^2$$$$\frac{x}{y}=u\Rightarrow (1-u^2)=5{(1-u)}^2$$$$(1-u)(1+u)-5{(1-u)}^2=0$$$$(1-u)(1+u-5+5u)=0$$$$(1-u)(6u-4)=0$$$$u=1\mid u=\frac{2}{3}$$$$x:y=1:1\mid x:y=2:3$$

Answered by MJS_new last updated on 09/Apr/21

$$(y-x)(y+x)=5{(y-x)}^2$$$$\left(1\right)y-x=0\Rightarrow \frac{x}{y}=1$$$$\left(2\right)y+x=5(y-x)\Rightarrow \frac{x}{y}=\frac{2}{3}$$

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