Answer
$\begin{bmatrix}13\\7\end{bmatrix},\begin{bmatrix}2x_{1}-x_{2}\\5x_{1}+6x_{2}\end{bmatrix}$
Work Step by Step
The key is to notice that the matrix of a linear transformation $T: \mathbb{R}^{n} \rightarrow \mathbb{R}^{m}$ is always of the form $\begin{bmatrix}T(\mathbf{e_{1}})&T(\mathbf{e_{2}})&\cdots &T(\mathbf{e_{n}})\end{bmatrix}$, where each $\mathbf{e_{i}}$ is the $i$th column of the $m\times m$ identity matrix, and $T(\mathbf{e_{i}})$ is its image under $T$. (This takes some getting used to. Take time to understand the statement, and test it using a few different matrices to see why it works.)
Because of this, the matrix of the given linear transformation is simply $\begin{bmatrix}T(\mathbf{e_{1}})&T(\mathbf{e_{2}})\end{bmatrix}=\begin{bmatrix}\mathbf{y_{1}}&\mathbf{y_{2}}\end{bmatrix}=\begin{bmatrix}2&-1\\5&6\end{bmatrix}$. Therefore, we can find the required images via multiplication of the given vectors by this matrix:
$\begin{bmatrix}2&-1\\5&6\end{bmatrix}\begin{bmatrix}5\\-3\end{bmatrix}=\begin{bmatrix}(2)(5)+(-1)(-3)\\(5)(5)+(6)(-3)\end{bmatrix}$
$\begin{bmatrix}2&-1\\5&6\end{bmatrix}\begin{bmatrix}x_{1}\\x_{2}\end{bmatrix}=\begin{bmatrix}(2)(x_{1})+(-1)(x_{2})\\(5)(x_{1})+(6)(x_{2})\end{bmatrix}$
