Integer solutions of $2x+3y=n$ What is the smallest $m$ such that for all $n\geq m$, the equation $2x+3y=n$ has solutions with $x,y \in \mathbb{Z}$ and $x,y\geq2$?
My approach. We can write the solutions in terms of the parameter $t$ as:
$$x(t)=-n-3t$$ and 
$$y(t) = n+2t$$
Setting both of those greater than or equal to $2$, we have $-n-3t\geq2$ and $n+2t\geq2$. Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
$$-2n-4\geq6-3n$$
So $n\geq10$. This seems to be right. For $n=10$, take $(2,2)$. But, suppose $n=11$. Then, $2x+3y=11$ clearly has no solution where both $x,y\geq2$. Is my solution incorrect? Can someone point me in the right direction? 
EDIT: Thanks for the great answers!

You are almost there.

Solving for $t$, we find $\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}$. Solving the resulting inequality, we get
  $$-2n-4\geq6-3n$$
  So $n\geq10$.

You want to find $n$ such that there is at least one integer $t$ satisfying
$$\dfrac{-n-2}{3} \geq t \geq\dfrac{2-n}{2}\tag1$$
For this, solving $\frac{-n-2}{3}\ge\frac{2-n}{2},$ i.e. $-2n-4\ge 6-3n$ is not enough.
If $\frac{-n-2}{3}-\frac{2-n}{2}\ge 1$, i.e. $n\ge 16$, then $(1)$ has at least one integer $t$.
For $n=11$, there is no $(x,y)$.
$2\cdot 3+3\cdot 2=12$.
$2\cdot 2+3\cdot 3=13$.
$2\cdot 4+3\cdot 2=14$.
$2\cdot 3+3\cdot 3=15$.
So, the answer is $12$.
