Find the number of positive integers solutions of the equation $3x+2y=37$

Find the number of positive integers solutions of the equation $3x+2y=37$ where $x>0,y>0,\ \ x,y\in \mathbb{Z}$ .

By trial and error I found

$$\begin{array}{|c|c|} \hline
x & y \\ \hline
11 & 2 \\ \hline
9 & 5 \\ \hline
7 & 8 \\ \hline
5 & 11 \\ \hline
3 & 14 \\ \hline
1 & 17 \\ \hline
\end{array}$$

Total $\large 6$ pair of solutions.
But i would like to know if their is a specific method to only find the number of positive solutions and not necessarily the actual solutions.

I look for a short and simple way.
I have studied maths upto $12th$ grade.

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In the world of combinatorics, you can use something commonly known as the “stars and bars” method. Generally put, the equation $$x_1+x_2+\ldots+ x_k = s$$ where $s,x_i$ are positive integers has $\binom{s-1}{k-1}$ many solutions. The quantity $\binom{s-1}{k-1}$ is called a binomial coefficient. A general binomial coefficient $\binom{a}{b}$ where $a,b$ are nonnegative integers with $a \geq b$ is defined as $$\binom{a}{b} =\frac{a!}{b!(a-b)!}$$ which means $$ \binom{s-1}{k-1} = \frac{(s-1)!}{(k-1)!(s-k)!}$$ In your problem we have $s = 37$ and $k = 2$. So, we want to calculate $$\binom{37-1}{2-1} = \frac{36!}{1!\space 35!} = 36$$ However, this tells us that the equation $x+y = 37$ has $36$ solutions. Since your equation is $3x+2y = 37$, we’ll want every third $x$ and every second $y$. This is tantamount to dividing $36$ by $3$ and $2$. We get $\frac{36}{3\cdot 2} = 6$, which matches your calculation. (Had the ratio not been an integer, you’d want to round down to the nearest integer). Depending on how large $k$ and $s$ are, you may find the stars and bars method to be preferable to calculating every solution by hand. It really boils down to a pretty easy factorial calculation. I’d say your problem could go either way.

Take any of your six positive solutions, say the first one $(11, 2)$; you have $3\cdot11 +2\cdot2 = 37$ and for another solution $(x, y)$ you have $3x +2y = 37$.

Hence $3(x-11)+2(y-2)=0$ and all the solutions are given by $(x, y) = (11 +2t, 2-3t)$. Do you want $11 + 2t \gt 0$ and $2 – 3t \gt 0$ therefore $-5.5 \lt t \lt 1$.

Thus the only positive integer solutions correspond to the parameter $t = -5, -4, -3, -2 , -1, 0$ exactly the six ones you have found.

Once you have found a duet of numbers $(x,y)$ satisfying this equation, let consider another duet as
$$
(x_1=x+a, y_1=y-b)
$$
For this second duet to satisfy the same equation, it is needed that

$$
3a-2b=0
$$

Which leads to all the possible duets written as
$$
(x+2n, y-3n)
$$
With $n$ in $\mathbb{N}$.