Do the following system of equation have a solution in $\mathbb{Z}_{p}$

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Consider the equations $5x + 3y = 4$ and $3x + 6y = 1$. I need to find if these equations has a solution in $\mathbb{Z}_{3}$ or $\mathbb{Z}_{7}$.

I solved these equations in the usual way and found that $x = 1$ and $y = -\frac{1}{3}$. I know that $\mathbb{Z}_{p}$ contains numbers from $0$ to $p-1$. How do I check if the above $x$ and $y$ are in $\mathbb{Z}_{3}$ or $\mathbb{Z}_{7}$.?

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As $x=1,y=-\dfrac13=(-3)^{-1}$ satisfy both

but in $\pmod3$ or $Z_3,$ there is no inverse of $-3$

In $Z_7,$ $$(-3)^{-1}\equiv2\pmod7$$

Alternatively for obvious reason, $3(x+2y)\equiv1\pmod3$ is untenable right?