How many solutions to this linear diaphontine equation ?

• The least values of $(x,y)$ should be $(1,1)$ . So we can let $x = p+1$ and $y = q+1$ . So now we can reframe $p,q$ be non-negative and the problem to:

$3(p+1)+ (q+1) \leq 6 \quad \implies \quad 3p+q \leq 2$

• Now it is obvious to see that $p$ has to be 0, hence we are left to check non-negative integers $q$ for

$q \leq 2 \quad \implies \quad q \in \{0, 1, 2\}$

• Hence we get three total cases. In terms of $(p,q)$ they are $(0,0), (0,1), (0,2)$ . They can be transformed to in terms of $(x,y)$ by adding 1 to each od the entries. $(1,1), (1,2), (1,3) \quad \implies \quad \boxed{3}$

$y\leq 3(2-x)$ , since both $x$ and $y$ are positive integers, therefore the only permitted value of $x$ is $1$ , and the only possible pairs are determined by $y\leq 3$ , the pairs are $(x, y) =(1,1),(1,2),(1,3)$ . So, there are $\boxed 3$ pairs.

$x,y>0, 3x+y\leq 6$ $\begin{array}{l|c}x&y\\\hline\\1&1,2,3,{\cancel{\color{#D61F06}{4}}(3x+y\leq 6)}\\{\cancel{\color{#D61F06}{2}}}&{\cancel{\color{#D61F06}{0}}(>0)}\end{array}$ $\text{Total pairs of }(x,y) =\boxed{3}$

Equations of these types are known as Diaphontine's eqution. Diaphontine's equation wiki link Since x and y are linear these are called as linear diaphontine's equation. The possible values of (x,y) which satisfy the equation 3x+ y <= 6 given that x and y are integers, are (1,1) (1,2) and (1,3). So there are 3 solutions to this linear diaphontine's equation.