Answer
$y=v_4-v_2-\frac{1}{2}v_3$
$[5\ 3]=[3\ 7]-[-2\ 2]-\frac{1}{2}[0\ 4]$
Work Step by Step
In this problem, we want to present $y=[5\ 3]$ as a linear combination of the basis vectors $v_1=[1\ 2]$, $v_2=[-2\ 2]$, $v_3=[0\ 4]$, and $v_4=[3\ 7]$, if possible. Through a trial and error process in which we obtain the desired first value and then counterbalance by an adjustment in the second, we find $y=v_4-v_2-\frac{1}{2}v_3$.
