Let $a=\pmatrix{1\\1}, b=\pmatrix{1\\2}, c=\pmatrix{2\\1}$. Solve $xa + yb = c$ for $x, y \in \mathbb{C}$.
Does this mean I do $1x+1y=2$ $1x+2y=1$ and would be the final answer?
Let $a=\pmatrix{1\\1}, b=\pmatrix{1\\2}, c=\pmatrix{2\\1}$. Solve $xa + yb = c$ for $x, y \in \mathbb{C}$.
Does this mean I do $1x+1y=2$ $1x+2y=1$ and would be the final answer?
Assuming Zev's edit is correct, then your equations $1x+1y=2$ $1x+2y=1$ are correct. To complete the problem, you need to find the values of $x$ and $y$ that make both of those equations true at the same time.
Hint: Subtract the first equation from the second equation.
$1x+1y=2\cdots\cdots(1)$ $1x+2y=1\cdots\cdots(2)$
Multiply (1) by -1 $-1x-1y=-2 $ Adding above and (2),
$1x+2y -1x-1y=1-2$ $y=-1$ Put it in (1) $1x+1y=2$ $1x-1=2$ $x=3$ Solution Set = (3,-1)
Verify this by putting in the 2 equations.