How do you solve $\dfrac{2}{3}x-5=-9$?

Hint: We separate the variables and the constants of the equation $\dfrac{2}{3}x-5=-9$. We apply the binary operation of addition and subtraction for the constants. The solutions of the variables and the constants will be added at the end to get the final answer to equate with 0. Then we solve the linear equation to find the value of $x$.

Complete step by step solution:
The given equation $\dfrac{2}{3}x-5=-9$ 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 $\dfrac{2}{3}x-5=-9$ are either variable of $x$ or a constant.
We take the constants all together to solve it. We add 5 on both of the equations.
$\begin{align}
  & \dfrac{2}{3}x-5+5=-9+5 \\
 & \Rightarrow \dfrac{2}{3}x=-4 \\
\end{align}$
Now we multiply both sides of the equation with $\dfrac{3}{2}$.
Therefore, $\dfrac{3}{2}\times \dfrac{2}{3}x=\dfrac{3}{2}\times \left( -4 \right)$. The simplified form is $x=-\dfrac{3\times 4}{2}=-\dfrac{12}{2}$.
For any fraction $\dfrac{p}{q}$, we first find the G.C.D of the denominator and the numerator. If it’s 1 then it’s already in its simplified form and if the G.C.D of the denominator and the numerator is any other number d then we need to divide the denominator and the numerator with d and get the simplified fraction form as $\dfrac{{}^{p}/{}_{d}}{{}^{q}/{}_{d}}$.
For our given fraction $\dfrac{12}{2}$, the G.C.D of the denominator and the numerator is 2.
Now we divide both the denominator and the numerator with 2 and get $\dfrac{{}^{12}/{}_{2}}{{}^{2}/{}_{2}}=\dfrac{6}{1}=6$.
The solution is \[x=-6\].

Note: To confirm the result of the solution being \[x=-6\] for equation $\dfrac{2}{3}x-5=-9$, we put the value on left side of the equation and find the final value.
Therefore, the left-hand side of the equation $\dfrac{2}{3}x-5=-9$ becomes
$\dfrac{2}{3}x-5=\dfrac{2}{3}\times \left( -6 \right)-5=-4-5=-9$
Thus, verified for the equation $\dfrac{2}{3}x-5=-9$ the solution is \[x=-6\].