The solution of linear inequalities $x+y\ge 5$ and $x-y\le 3$ lies:

Consider the inequalities:$x+y\ge 5$: The boundary line is $x+y=5$.

For $(0,0)$, $0\ge 5$ is false, so the region is away from the origin.$x-y\le 3$: The boundary line is $x-y=3$. For $(0,0)$, $0\le 3$ is true, so the region includes the origin.The intersection of these regions:The line $x+y=5$ passes through $(5,0)$ and $(0,5)$.The line $x-y=3$ passes through $(3,0)$ and $(0,-3)$.The intersection point is $(4,1)$.The feasible region extends into the first quadrant (where both $x,y$ can be positive) and continues into the second quadrant (as $x$ decreases and $y$ increases, satisfying $x+y\ge 5$ and $x-y\le 3$).Therefore, the solution lies in the first and second quadrants.