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.