Answer
See below
Work Step by Step
Given the set of vectors $\{(1,-5),(6,3)\}$
Obtain: $\begin{vmatrix}
1 & -5\\ 6 & 3
\end{vmatrix}=3+30=33 \ne 0$
Thus, the set of vectors are notcoplanar and therefore set $\{(1,-5),(6,3)\}$ does span $R^2$
Rewrite $v=(x,y)=a(1,-5)+b(6,3)$
We have the system:
$a+6b=x\\
-5a+3b=y$
then $5(a+6b)+(-5a+3b)=33b=5x+y\ rightarrow b=\frac{55x+y}{33}$
Substitute: $a=-6b+x=-6(\frac{55x+y}{33})+x=\frac{x-2y}{11}$
Hence vector $v=(5,-7)$ can be written as a linear combination of vectors $\{\frac{x-2y}{11}(1,-5)+\frac{5x+y}{33}(6,3)\}$
