Inequations $3 x-y \geq 3$ and $4 x-y>4$

Have solution for positive $x$ and $y$

Following figure will be obtained on drawing the graphs of given inequations :

From $3 x-y \geq 3, \frac{x}{1}+\frac{y}{-3}=1$

From $4 x-y \geq 4, \frac{x}{1}+\frac{y}{-4}=1$

Clearly the common region of both the inequations is true for positive value of (x, y). It is also true for positive values of x and negative values of y.