Answer:
[tex]\mathrm{The\:solution\:is}[/tex] :
[tex]x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}[/tex]
Step-by-step explanation:
Given
[tex]y=\sqrt{\frac{2x+1}{x-1}}[/tex]
Taking square of both sides
[tex]y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)\:^2[/tex]
[tex]\mathrm{Subtract\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2\mathrm{\:from\:both\:sides}[/tex]
[tex]y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=0[/tex]
As we know that [tex]\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)[/tex]
[tex]\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2x+1}{x-1}}\right)^2:\quad \left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)[/tex]
so
[tex]\left(y+\sqrt{\frac{2x+1}{x-1}}\right)\left(y-\sqrt{\frac{2x+1}{x-1}}\right)=0[/tex]
[tex]\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
[tex]\mathrm{Solve\:}\:y+\sqrt{\frac{2x+1}{x-1}}=0[/tex]
[tex]\mathrm{Subtract\:}y\mathrm{\:from\:both\:sides}[/tex]
[tex]y+\sqrt{\frac{2x+1}{x-1}}-y=0-y[/tex]
[tex]\sqrt{\frac{2x+1}{x-1}}=-y[/tex]
[tex]\mathrm{Square\:both\:sides}[/tex]
[tex]\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2[/tex]
[tex]\mathrm{Expand\:}\left(\sqrt{\frac{2x+1}{x-1}}\right)^2[/tex]
[tex]\left(\sqrt{\frac{2x+1}{x-1}}\right)^2[/tex]
[tex]\mathrm{Apply\:radical\:rule}:\quad \sqrt{a}=a^{\frac{1}{2}}[/tex]
[tex]=\left(\left(\frac{2x+1}{x-1}\right)^{\frac{1}{2}}\right)^2[/tex]
[tex]=\frac{2x+1}{x-1}[/tex]
so equation [tex]\left(\sqrt{\frac{2x+1}{x-1}}\right)^2=\left(-y\right)^2[/tex] becomes
[tex]\frac{2x+1}{x-1}=y^2[/tex]
now
[tex]\mathrm{Solve\:}\:\frac{2x+1}{x-1}=y^2[/tex]
[tex]\frac{2x+1}{x-1}=y^2[/tex]
[tex]\mathrm{Multiply\:both\:sides\:by\:}x-1[/tex]
[tex]\frac{2x+1}{x-1}\left(x-1\right)=y^2\left(x-1\right)[/tex]
[tex]2x+1=y^2\left(x-1\right)[/tex]
[tex]2x+1=xy^2-y^2[/tex] ∵ [tex]y^2\left(x-1\right):\quad xy^2-y^2[/tex]
[tex]2x=xy^2-y^2-1[/tex]
[tex]2x-xy^2=-y^2-1[/tex]
[tex]x\left(2-y^2\right)=-y^2-1[/tex] ∵ [tex]\mathrm{Factor}\:2x-xy^2:\quad x\left(2-y^2\right)[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}2-y^2[/tex]
[tex]\frac{x\left(2-y^2\right)}{2-y^2}=-\frac{y^2}{2-y^2}-\frac{1}{2-y^2}[/tex]
[tex]x=\frac{-y^2-1}{2-y^2}[/tex]
so
[tex]y+\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\le \:0\right\}[/tex]
similarly
[tex]y-\sqrt{\frac{2x+1}{x-1}}=0:\quad x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}[/tex]
[tex]\mathrm{Verify\:Solutions}:\quad x=\frac{-y^2-1}{2-y^2}[/tex]
[tex]\mathrm{Check\:the\:solutions\:by\:plugging\:them\:into\:}y^2=\left(\sqrt{\frac{2x+1}{x-1}}\right)^2[/tex]
[tex]\mathrm{Remove\:the\:ones\:that\:don't\:agree\:with\:the\:equation.}[/tex]
[tex]\mathrm{Plug}\quad x=\frac{-y^2-1}{2-y^2}[/tex]
[tex]y^2=\left(\sqrt{\frac{2\left(\frac{-y^2-1}{2-y^2}\right)+1}{\left(\frac{-y^2-1}{2-y^2}\right)-1}}\right)^2[/tex]
[tex]\mathrm{Subtract\:}\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2\mathrm{\:from\:both\:sides}[/tex]
[tex]y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2[/tex]
[tex]y^2-\left(\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2=0[/tex]
[tex]\mathrm{Factor\:}y^2-\left(\sqrt{\frac{2\frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)^2:\quad \left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)[/tex]
so
[tex]\left(y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)\left(y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}\right)=0[/tex]
[tex]\mathrm{Using\:the\:Zero\:Factor\:Principle:\quad \:If}\:ab=0\:\mathrm{then}\:a=0\:\mathrm{or}\:b=0\:\left(\mathrm{or\:both}\:a=0\:\mathrm{and}\:b=0\right)[/tex]
[tex]\mathrm{Solve\:}\:y+\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\le \:0[/tex]
[tex]\mathrm{Solve\:}\:y-\sqrt{\frac{2\cdot \frac{-y^2-1}{2-y^2}+1}{\frac{-y^2-1}{2-y^2}-1}}=0:\quad y\ge \:0[/tex]
[tex]\mathrm{True\:for\:all}\:y[/tex]
Therefore, [tex]\mathrm{The\:solution\:is}[/tex] :
[tex]x=\frac{-y^2-1}{2-y^2}\space\left\{y\ge \:0\right\}[/tex]
