They are equal, because the extra zero adds nothing.

They are different vectors, living in $\mathbb{R}^2$ and $\mathbb{R}^3$.

They are equal after both are placed in $\mathbb{R}^3$.

Only the second is a vector.

$(2,4)$

$(5,1)$

$(2,-2,0,6)$

$(3,1)$

$(4,2)$

$(2,4,0)$

$(-1,-2)$

$(1,3)$

The whole of $\mathbb{R}^2$.

The two given vectors only.

The set $\{(0,0)\}$.

The line of all vectors with second component zero.

$4x-\tfrac{1}{2}y=9$

$xy+x=9$

$x^2-y^2=9$

$\sqrt{x}+\sqrt{y}=9$

$(3,-1)$

$(5,-1)$

$(1,5)$

$(2,0)$

$x+y=13$

$x-2y=13$

$x+y=5$

$x-5y=1$

It has exactly one solution.

It has infinitely many solutions.

It has no solution.

Nothing can be said without eliminating first.