Set the equations equal: $2x+3=-x+9 \Rightarrow 3x=6 \Rightarrow x=2$. Then $y=2(2)+3=7$. Check: In $y=2x+3$, $y=7$; in $y=-x+9$, $y=-2+9=7$. So $(2,7)$ satisfies both. The intersection is the $(x,y)$ that makes both equations true.

Both lines have the same slope $m=3$ but different $y$-intercepts ($-4$ and $1$), so they are parallel and never meet. No $(x,y)$ satisfies both at once. Therefore, there is no solution (no intersection).

Set equal: $\tfrac{1}{2}x-1=-2x+5 \Rightarrow \left(\tfrac{1}{2}+2\right)x=6 \Rightarrow \tfrac{5}{2}x=6 \Rightarrow x=\tfrac{12}{5}$. Then $y=\tfrac{1}{2}\cdot\tfrac{12}{5}-1=\tfrac{6}{5}-1=\tfrac{1}{5}$. Check in $y=-2x+5$: $-2\cdot\tfrac{12}{5}+5=-\tfrac{24}{5}+\tfrac{25}{5}=\tfrac{1}{5}$. The intersection is $(\tfrac{12}{5},\tfrac{1}{5})$. Intersection means the $(x,y)$ that satisfies both equations.

Both equations represent the same line. Every point on $y=-x+4$ satisfies both equations, so there are infinitely many solutions. Intersection here means all $(x,y)$ on that line satisfy both.

Set equal: $-3x+2=x-6 \Rightarrow -4x=-8 \Rightarrow x=2$. Then $y=-3(2)+2=-4$. Check in $y=x-6$: $2-6=-4$. So $(2,-4)$ satisfies both. The intersection is the ordered pair making both equations true.