Number of non-negative integer solutions of an equation using generating functions Using convolution, find how many non-negative integer solutions of
$$\begin{aligned} x_1 + 5 x_2 &= n\\ x_1 + 5 x_2 &= 60\end{aligned}$$
Is anyone able to solve this problem or figure out the idea? Thank you so much.

Here's a combinatorial interpretation: the number of solutions to the second equation is the number of ways to make 60 cents in change using 1 cent and 5 cent coins.
I assume that the second equation is a special case of the first, and that you want to know how to solve both.  The method for $n=60$ illustrates the general case.
For a non-negative integer $a$, let $r_a$ be the number of non-negative integer solutions to $x_1=a$ and let $s_a$ be the number of non-negative integer solutions to $5x_2=a$.  So $r_a$ is the number of ways to make $a$ cents in change using 1 cent coins alone, and $s_a$ is the number of ways of making $a$ cents in change using 5 cent coins alone.  Can you characterize $r_a$ and $s_a$?  The number of non-negative integer solutions to $x_1+5x_2=60$ is then given by the convolution
$$r_0s_{60}+r_1s_{59}+\ldots+r_{60}s_0.$$
This is the coefficient of $x^{60}$ in the product of the generating functions for the coefficients $r_a$ and $s_a,$
$$(r_0+r_1x+r_2x^2+\ldots)(s_0+s_1x+s_2x^2+\ldots).$$
You should have found that $r_a=1$ for all $a$ and that $s_a=1$ for $a$ divisible by 5 and $s_a=0$ otherwise.  Then
$$r_0+r_1x+r_2x^2+\ldots=1+x+x^2+\ldots=\frac{1}{1-x}$$
and
$$s_0+s_1x+s_2x^2+\ldots=1+x^5+x^{10}+\ldots=\frac{1}{1-x^5}.$$
Here's one method for extracting the coefficient of $x^n$ in the generating function
$$\frac{1}{(1-x)(1-x^5)}.$$
Write
$$\frac{1}{1-x}=\frac{1+x+x^2+x^3+x^4}{1-x^5}.$$
Then
$$\frac{1}{(1-x)(1-x^5)}=\frac{1+x+x^2+x^3+x^4}{(1-x^5)^2}=(1+x+x^2+x^3+x^4)\sum_{r=0}^\infty\binom{-2}{r}(-x^5)^r,$$
where $\binom{-2}{r}=\frac{(-2)(-3)\ldots(-2-r+1)}{r!}$ is the generalized binomial coefficient.  You can show that $(-1)^r\binom{-2}{r}=r+1$, and so the generating function is
$$(1+x+x^2+x^3+x^4)\sum_{r=0}^\infty(r+1)x^{5r}.$$
The $x^{60}$ term comes from multiplying the $1=x^0$ term in the polynomial with the $r=12$ term in the summation.  Hence the answer is $1\cdot(12+1)=13$.
This method can be generalized to the case where you have additional coin values, say 10 cent coins.
