Algebra to 3

Given that $4x+7y = 3 \implies x = \dfrac {3-7y}4 = \dfrac {3(1-y)}4 - y$ . For $x$ to be an integer, $1-y \equiv 0 \text{ (mod 4)} \implies y \equiv 1 \text{ (mod 4)}$ or $y = 4n+1$ , where $n$ is an integer. Then $x = - 7n - 1$ and

$\begin{aligned} - 100 < - 7x & - 1 < 100 \\ -99 < - 7n & < 101 \\ -101 < 7n & < 99 \\ -14 \le n & \le 14 \end{aligned}$

Therefore there are $29$ $n$ 's and hence $\boxed{29}$ integer pair $(x,y)$ solutions.