Solve the following for x:
$\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.

Hint: Subtract $\dfrac{1}{2a}$ from both the sides of the provided equation. Now, take L.C.M on both sides and simplify the equation. Take the common terms together by taking all the terms to L.H.S. and write the overall equation as a product of several terms containing ‘x’. Substitute each term, containing ‘x’, equal to 0 and find the value of x.

Complete step-by-step answer:
We have been provided with the equation: $\dfrac{1}{2a+b+2x}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}$.
Now, subtracting $\dfrac{1}{2a}$ from both the sides of the equation, we get,
$\begin{align}
  & \dfrac{1}{2a+b+2x}-\dfrac{1}{2a}=\dfrac{1}{2a}+\dfrac{1}{b}+\dfrac{1}{2x}-\dfrac{1}{2a} \\
 & \Rightarrow \dfrac{1}{2a+b+2x}-\dfrac{1}{2a}=\dfrac{1}{b}+\dfrac{1}{2x} \\
\end{align}$
Taking L.C.M on both sides, we get,
$\begin{align}
  & \dfrac{2a-\left( 2a+b+2x \right)}{\left( 2a+b+2x \right)\left( 2a \right)}=\dfrac{2x+b}{2bx} \\
 & \Rightarrow \dfrac{-\left( b+2x \right)}{\left( 2a+b+2x \right)\left( 2a \right)}=\dfrac{b+2x}{2bx} \\
\end{align}$
Taking all the terms to the L.H.S, we get,
$\begin{align}
  & \dfrac{-\left( b+2x \right)}{\left( 2a+b+2x \right)\left( 2a \right)}-\dfrac{\left( b+2x \right)}{2bx}=0 \\
 & \Rightarrow -\left( \dfrac{\left( b+2x \right)}{\left( 2a+b+2x \right)\left( 2a \right)}+\dfrac{\left( b+2x \right)}{2bx} \right)=0 \\
\end{align}$
Multiplying both the sides with (-1), we get,
$\dfrac{\left( b+2x \right)}{\left( 2a+b+2x \right)\left( 2a \right)}+\dfrac{\left( b+2x \right)}{2bx}=0$
Taking the numerator common, we get,
$\begin{align}
  & \left( b+2x \right)\left( \dfrac{1}{\left( 2a+b+2x \right)\left( 2a \right)}+\dfrac{1}{2bx} \right)=0 \\
 & \left( b+2x \right)\left( \dfrac{1}{\left( 2a+b+2x \right)a}+\dfrac{1}{bx} \right)=0 \\
\end{align}$
Taking L.C.M on the L.H.S, we get,
\[\begin{align}
  & \left( b+2x \right)\left( \dfrac{bx+\left( 2a+b+2x \right)a}{\left( 2a+b+2x \right)a} \right)=0 \\
 & \Rightarrow \left( b+2x \right)\left( \dfrac{bx+2{{a}^{2}}+ab+2ax}{\left( 2a+b+2x \right)a} \right)=0 \\
 & \Rightarrow \left( b+2x \right)\left( \dfrac{\left( 2{{a}^{2}}+ab \right)+\left( 2ax+bx \right)}{\left( 2a+b+2x \right)a} \right)=0 \\
 & \Rightarrow \left( b+2x \right)\left( \dfrac{a\left( 2a+b \right)+x\left( 2a+b \right)}{\left( 2a+b+2x \right)a} \right)=0 \\
 & \Rightarrow \left( b+2x \right)\left( \dfrac{\left( 2a+b \right)\left( a+x \right)}{\left( 2a+b+2x \right)a} \right)=0 \\
\end{align}\]
Now, multiplying both sides with \[\left( 2a+b+2x \right)a\], we get,
\[\left( b+2x \right)\left( 2a+b \right)\left( a+x \right)=0\]
Dividing both sides by (2a + b), we have,
\[\left( b+2x \right)\left( a+x \right)=0\]
Substituting the terms, containing ‘x’, equal to 0, we get,
\[\begin{align}
  & \left( a+x \right)=0\text{ or }\left( b+2x \right)=0 \\
 & \Rightarrow x=-a\text{ or }2x=-b \\
 & \Rightarrow x=-a\text{ or }x=\dfrac{-b}{2} \\
\end{align}\]

Note: One may note that we can also solve this question by directly taking the L.C.M in R.H.S without subtracting $\dfrac{1}{2a}$ from each side. The next step will be cross-multiplication and then taking the common terms together. Finally substituting all the terms containing ‘x’ equal to 0, we will get the answer. But the main problem we will face in this process is that we will get so many terms on cross-multiplication, which may be confusing while we are grouping them.