Solve the equation $3x=2$ in the fields $\mathbb Z_7$ and $\mathbb Z_{23}$

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This is a sum from Abstract Algebra by Fraleigh.

Myy attempt:

$$3x=2$$ $$\Rightarrow 3x-2=0$$

Now, the elements of $\mathbb Z_7$ are {$0,1,2,3,4,5,6$}

Substituting these values in the left side of the equation, for $3$, $3\times 3-2= 7=0$ so $3$ is a solution. No other value satisfies the equation.

The elements of $\mathbb Z_{23}$ are {$0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22$}

Now, $3\times {16} -2= 46=0$, so $16$ is a solution. No other value satisfies the equation.

Is my method correct? Did I get all the solutions or did I miss out some?

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It is a brute forse method. For correct solution you have to find (by extended Euclide algorithm) the muliplicative inverse for $3$ and then multiply it by 2.