Prove $\begin{pmatrix} a&b\\ 2a&2b\\ \end{pmatrix} \begin{pmatrix} x\\y\\ \end{pmatrix}=\begin{pmatrix} c\\2c \\ \end{pmatrix}$ Prove that
$$
\begin{pmatrix} a & b \\ 2a & 2b \\ \end{pmatrix}
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} c \\ 2c \\ \end{pmatrix}
$$
has solution
$$
\begin{pmatrix} x \\ y \\ \end{pmatrix}
=
\begin{pmatrix} \frac{c}{a} \\ 0 \\ \end{pmatrix}
+t
\begin{pmatrix} -b \\ a \\ \end{pmatrix}
$$
and $t$ is real number.
If I plug some numbers, all $y$ variables are vanish and the matrix has infinitely many solutions, but I have no idea how the solution could be into like that. Also I haven't reached Vector chapter.
Any help is appreciated!

The fast approach, as I see it is (in case $a \neq 0$):
you have $ax+by = c$ , just plug in $y=0+ta$, and you get 
$$ax + bta = c$$
$$x= \frac ca - bt$$. 
alternatively, in case you don't have the solution in advance, use same approach and you will get: $$ x = \frac ca - \frac ba y$$, so the general solution will be $$(x,y) = (\frac ca - \frac bat , t)$$ 
$$ (x,y) = (\frac ca, 0) + (-\frac ba,1)t$$ where $t$ is free. if you plug $t=a$ you will end up with the same result. 
