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What would be the best method to solve the systems of equations? If solving it, please show how to solve for $x$ and $y$ because that's where it's confusing. Thanks.

$5y^3+4x^2-y^2+10x-y$ $=$ $1$ $\pmod {31}$

$2y^3+13x^3+7y^2-x^2+5x-2$ $=$ $1$ $\pmod {31}$

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    A computer program might be easiest: there are only $31^2=961$ pairs $(x,y)$ to check, each can be plugged into the two equations and a success when both are true. – 2017-02-05
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    To solve $$\begin{split} P_1(x,y)&=0\\ P_2(x,y)&=0\end{split}$$ for $x$ and $y$, start with eliminating the highest power of, say, $y$. In this case both equations have an $y^3$ term, but in general you use the equations where the highest power of $y$ is the lowest. If that highest power is $y^n$, then use that equation to express $y^n$ in terms of the lower powers of $y$ and $x$. Then multiply this by powers of $y$ to express higher powers of $y$ that occur in the other equation in terms of lower powers of $y$. Then substitute these relations in the other equation. – 2017-02-05
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    You then end up with an equation where the highest power of $y$ is $y^{n-1}$ or lower. You then play the same game again, this time using that equation to eliminate $y^n$ and $y^{n-1}$ from the equation you started with. Clearly, this process will eventually eliminate $y$ completely, you'll end up with a polynomial equation containing only $x$. – 2017-02-05

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Systematically eliminating $y$ by starting with the cubic term and then moving on to lower powers of $y$ leads to the equation:

$$16 x^9+9 x^8+10 x^7+14 x^6+28 x^5+28 x^4+19 x^3+13 x^2+2 x+26=0 \bmod 31$$

and

$$y = \frac{15 x^3+11 x^2+29 x+9}{x^3+6 x^2+12 x+15}\bmod 31$$

where the denominator is to be interpreted as the multiplicative inverse of the polynomial $\bmod 31$ there, which can be rewritten as a polynomial (because it doesn't have linear factors), but we don't need to do that. Then we can solve for $x$ using trial and error, we find that $x=11\bmod 31$, substituting in the equation for $y$ yields $y = 16\bmod 31$.