How do you solve $\dfrac{4}{u+6}=\dfrac{6}{u+6}+2$?

Hint: First of all to solve the above equation we have to simplify the given equation into simplified form and then we have to do the calculation. Certain transformations and certain substitutions are to be made to get the value of $u$. After getting the value of $u$ we have to verify both sides of the equation to make sure whether we got the exact value or not.

Complete step by step answer:
From the question it had been given that, $\dfrac{4}{u+6}=\dfrac{6}{u+6}+2$
First of all we need to put the requirements, that is $u\ne -6$ because then the denominator will be zero and make the equation undefined.
$\dfrac{4}{u+6}=\dfrac{6}{u+6}+2$
On further simplification, we get
$\dfrac{4}{u+6}-\dfrac{6}{u+6}=2$
Since the denominators are the same, we can subtract 4 and 6 and we get,
$\Rightarrow \dfrac{-2}{u+6}=2$
Now, we will cross multiply and we will get,
$\Rightarrow -2=2u+12$
Keeping the variable u on one side and taking 12 to the LHS, we get
$\Rightarrow 2u=-14$
Dividing both sides by 2, we get
$\Rightarrow u=-7$

Therefore the value of $u=-7$.

Note: We have to do the verification process to check whether $u=-7$ is the exact value or not.
Verification:
Left hand side: first we have to substitute $u=-7$ in the left hand side of the equation.
On substituting $u=-7$ in left hand side of the equation we get,
$\begin{align}
  & \dfrac{4}{u+6}=\dfrac{4}{-7+6} \\
 & \Rightarrow -4 \\
\end{align}$
Right hand side: first we have to substitute $u=-7$ in the right hand side of the equation.
On substituting $u=-7$ in right hand side of the equation we get,
$\begin{align}
  & \dfrac{6}{u+6}+2=\dfrac{6}{-7+6}+2 \\
 & \Rightarrow -6+2 \\
 & \Rightarrow -4 \\
\end{align}$
Therefore, we can clearly observe that the left hand side of the equation is equal to the right hand side of the equation.
Hence, verified.