How do you solve $\dfrac{2a-4}{6}=\dfrac{4a}{3}+\dfrac{2}{3}$ ?

Hint: This is a linear equation in one variable, the equation is in fraction form; we can multiply 3 to both LHS and RHS to convert it into simple form. Then we can separate variables and constants and divide LHS and RHS by coefficient of a to find the value of a.

Complete step by step solution:
The given equation in the question is $\dfrac{2a-4}{6}=\dfrac{4a}{3}+\dfrac{2}{3}$
Multiplying 3 both sides we get a – 2 = 4a + 2
Now we can bring all the a to LHS and all the constant to RHS
Adding 2 to both LHS and RHS we get a = 4a + 4
Now subtracting 4a from LHS and RHS we get -3a = 4
We can see the coefficient of a is -3, so by dividing LHS and RHS by -3 we get a equal to $-\dfrac{4}{3}$
So a = $-\dfrac{4}{3}$ is the solution of $\dfrac{2a-4}{6}=\dfrac{4a}{3}+\dfrac{2}{3}$

Note: The equation given in the question was a linear equation in one variable, so one equation is enough to solve it. We can check whether our answer is correct or not by putting the value of a in the equation. We know that the equation $\dfrac{2a-4}{6}=\dfrac{4a}{3}+\dfrac{2}{3}$ is equal to the equation a – 2 = 4a + 2. So by putting $-\dfrac{4}{3}$ in a – 2 = 4a + 2 , we get $-\dfrac{4}{3}$ - 2 = $-\dfrac{16}{3}$ + 2 which is correct. So a = $-\dfrac{4}{3}$ is the correct solution of $\dfrac{2a-4}{6}=\dfrac{4a}{3}+\dfrac{2}{3}$