Well, first of all you should know that given a linear transformation $T : V \to W$ if you know the values of $T$ on the basis of $V$ then you know $T$ completely by virtue of linearity. The point here is: in an $n$-dimensional vector space, $n$ linearly independent vectors necessarily forms a basis.
In this case $\mathbb{R}^2$ has dimension $2$ so that any set of $2$ linearly independent vector is a basis for $\mathbb{R}^2$. It's easy to see that $v_1$ and $v_2$ are linearly independent because they're not multiple one of the other. Hence the set $\left\{v_1, v_2\right\}$ is a basis of $\mathbb{R}^2$.
Now here comes the point. Given any $(x,y) \in \mathbb{R}^2$, how do we write it in this new basis? Well, it's easy, it should be a linear combination, so that we must have:
$$(x,y) = av_1+bv_2$$
And substituting we get:
$$(x,y) = a(1, -1) + b(2, -3)$$
$$(x,y) = (a+2b, -a-3b)$$
This is a system of linear equations in $a$ and $b$. If you solve, you'll find that $a = 3x+2y$ and $b =-x-y$. So that any vector $(x,y) \in \mathbb{R}^2$ is given in that basis by:
$$(x,y) = (3x+2y)v_1 + (-x-y)v_2$$
And hence
$$T(x,y) = (3x+2y)T(v_1) + (-x-y)T(v_2)$$
Now it's just a question of substituting the values:
$$T(x,y) = (3x+2y)(7, -8) + (-x-y)(17,-19)$$
So that finally, the expression of $T$ on arbitrary $(x,y)\in \mathbb{R}^2$ is:
$$T(x,y) = (4x-3y, -5x+3y)$$
Which is really the expression of $T$. You can easily verify that $T(v_1)$ and $T(v_2)$ are what they are expected to be.
Best Answer
Note sure if
(homework)yet. So hint:Let $$ T = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$
We can re-interpret the given $T(v_1)$ and $T(v_2)$ as:
$$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 \\ -1 \end{pmatrix} = \begin{pmatrix} 15 \\ -6 \end{pmatrix} , \\ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -2 \\ -1 \end{pmatrix} = \begin{pmatrix} 11 \\ -3 \end{pmatrix} $$ Or more succinctly as, $$ \begin{pmatrix} a & b \\ c & d \end{pmatrix} \begin{pmatrix} -3 & -2 \\ -1 & -1 \end{pmatrix} = \begin{pmatrix} 15 & 11 \\ -6 & -3 \end{pmatrix} \tag{1} $$ Can you take it from here?