How do you solve $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$?

Hint: We separate the variable and the constants of the equation $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$. We apply the binary operation of addition for constants. We take the LCM form to complete the subtraction. The solutions of the variables and the constants will be inverted at the end to get the final answer. Then we solve the linear equation to find the value of $x$.

Complete step by step solution:
The given equation $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$ is a linear equation of $x$. We need to simplify the equation by solving the variables and the constants separately.
All the terms in the equation of $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$ are either variable of $x$ or a constant. We first separate the constants.
We take the constants all together to solve it.
$\begin{align}
  & \dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x} \\
 & \Rightarrow \dfrac{1}{x}=\dfrac{2}{3}+\dfrac{1}{5} \\
\end{align}$
The binary operation between the fractions is addition for which we need the LCM of the denominators.
The LCM of the numbers 3 and 5 is 15.
\[1\left| \!{\underline {\,
  3,5 \,}} \right. \]
So, we get $3\times 5=15$.
Now we complete the subtraction and get
$\Rightarrow \dfrac{2}{3}+\dfrac{1}{5}=\dfrac{5\times 2+1\times 3}{15}=\dfrac{13}{15}$
Therefore, the final form is $\dfrac{1}{x}=\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{13}{15}$.
Now we take the inverse of the equation in both sides to get $x=\dfrac{15}{13}$

The solution is $x=\dfrac{15}{13}$.

Note: We can verify the result of the equation $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$ by taking the equation as $\dfrac{2}{3}-\dfrac{1}{x}=-\dfrac{1}{5}$.
Therefore, the left-hand side of the equation $\dfrac{2}{3}-\dfrac{1}{x}=-\dfrac{1}{5}$ becomes
$\dfrac{2}{3}-\dfrac{1}{\dfrac{15}{13}}=\dfrac{2}{3}-\dfrac{13}{15}=\dfrac{5\times 2-13}{15}=-\dfrac{3}{15}=-\dfrac{1}{5}$
Thus, verified for the equation $\dfrac{2}{3}+\dfrac{1}{5}=\dfrac{1}{x}$ the solution is $x=\dfrac{15}{13}$.