Quick Algebra Identities Repeated

$\large\Rightarrow \dfrac{ \sqrt{2x-1} +\sqrt{2x+1} }{ \sqrt{2x+1} - \sqrt{2x-1} } = \dfrac{5}{3}$ Applying Componento and Dividendo , $\large \Rightarrow\dfrac{ (\sqrt{2x-1} +\sqrt{2x+1} ) +(\sqrt{2x+1} - \sqrt{2x-1})}{(\sqrt{2x-1} +\sqrt{2x+1}) - (\sqrt{2x+1} - \sqrt{2x-1} )} = \dfrac{5+3}{5-3}$ This simplifies to

$\large\Rightarrow \dfrac{\sqrt{2x+1}}{\sqrt{2x-1}}=4$ Squaring both sides and cross multiplying the terms, $\large\Rightarrow 2x+1+16(2x-1)$ $\large\Rightarrow x=\dfrac{17}{30}$

$\Rightarrow\dfrac{ \sqrt{2x-1} +\sqrt{2x+1} }{ \sqrt{2x+1} - \sqrt{2x-1} } = \dfrac{5}{3}$
Let $\sqrt{2x+1}=a$ and $\sqrt{2x-1}=b$ .
$\dfrac{b+a}{a-b}=\dfrac{5}{3}$
Multiplying by $(a+b)$ on numerator and denominator.
$\dfrac{b+a}{a-b}×\dfrac{a+b}{a+b}=\dfrac{5}{3}$
$\dfrac{a^2+b^2+2ab}{a^2-b^2}=\dfrac{5}{3}$
$\dfrac{2x+\sqrt{4x^2-1}}{1}=\dfrac{5}{3}$
Now cross multiplying.
$6x+3\sqrt{4x^2-1}=5$
$6x-5=-3\sqrt{4x^2-1}$
Squring both sides.
$36x^2+25-60x=36x^2-9$
$-60x=-34$
$x=\boxed{\dfrac{17}{30}}$

Let $\sqrt { 2x-1 } =A$ Let $\sqrt { 2x+1 } =B$

$\frac { A+B }{ B-A } =\frac { 5 }{ 3 }$ $3A+3B=5B-5A$ $8A=2B$ $4A=B$ $(4A)^{ 2 }=(B)^{ 2 }$ $32x-16=2x+1$ $30x=17$ $x=\frac { 17 }{ 30 }$

Hmmm... All the previous solutions look more complicated (I have no idea what Componento and Dividendo is lol). Here is my solution: $\Rightarrow \frac { a+b }{ b-a } =\frac { 5 }{ 3 }$ $\Rightarrow 5b-5a=3a+3b$ $\Rightarrow 2b=8a\Rightarrow b=4a$ $\Rightarrow \sqrt { 2x+1 } =4\sqrt { 2x-1 }$ Square both sides to yield: $\Rightarrow 2x+1=32x-16$ And so $\Rightarrow 17=30x$ $\Rightarrow x=\frac { 17 }{ 30 }$