To get from here
$\displaystyle \bigg[\begin{array}{cc}2&2\\-1&5\end{array}\bigg]\bigg[\begin{array}{c}v_1\\v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]$
to here
$\displaystyle -v_1+2v_2=0$
We do this:
First multiply the matrices together:
$\displaystyle \bigg[\begin{array}{cc}2&2\\-1&5\end{array}\bigg]\bigg[\begin{array}{c}v_1\\v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]=\bigg[\begin{array}{cc}2v_1&2v_2\\-1v_1&5v_2\end{array}\bigg]$
Now set the two matrices equal to each other:
$\displaystyle \bigg[\begin{array}{cc}2v_1&2v_2\\-1v_1&5v_2\end{array}\bigg]=\bigg[\begin{array}{c}3v_1\\3v_2\end{array}\bigg]$
We can write the matrices as a system of equations:
$\displaystyle \left\{\begin{array}{rcl} 2v_1+2v_2 & = & 3v_1\\-v_1+5v_2 & = & 3v_2\end{array}\right.$
Thus, if we set each equation equal to zero, we have
$\displaystyle \left\{\begin{array}{rcl} -v_1+2v_2 & = & 0\\-v_1+2v_2 & = & 0\end{array}\right.$
Now, how do we determine $\displaystyle v_1$ and $\displaystyle v_2$?
There are infinitely many solutions.
I'll pick a simple one: $\displaystyle v_1=1$ and $\displaystyle v_2=\tfrac{1}{2}$
These two values, and any integer multiple of them will satisfy the equation we came up with.
Does this make sense?
--Chris