2016/007) How many integer values of x and y are there such that $4x+7y=3$ while $ |x| < 500$ and $|y| < 500$

1st let us find one solution
this can be found by any method but as we see that $7- 4 = 3$
so $(-1,1)$ is a solution
as coeffcient of y is larger so we need to restrict x between $- 500$ to $500$
general solution is $x = -1+ 7t$ and $y = 1 + 4t$
Now $- 500 < x < 500$ or -$ 500 < 1 + 7t < 500$ or
$-501 < 7t < 499$ or $-71 <= t <= 71$ so 143 values