I have this two signals $$x[n]=4\delta[n+1]+\delta[n-2]-2\delta[n-5]$$ $$y[n]=(3n-5)(u[n-1]-u[n-4]$$
I know that $y[n]$ results by the linear transformation of $x[n]$ and I have to find the parameters that make this possible.
Since $x[-1]=y[3]$ and $x[2]=y[2]$, solving this system $\left\{\begin{matrix} x[-1]=x[3a-b]& \\ x[2]=x[2a-b]& \end{matrix}\right.$ I got that $a=-3$ and $b=-8$, so the linear transformation is characterized by a time shift of $-8$ and a time scaling of $-3$.
My question is how to do the other way around, that is, imagine I have $$x[n]=4\delta[n+15]+2\delta[n+5]-8\delta[n-3]$$ and apply a time scaling of $a=4$ and a time shift of $b=9$ and I need to find the signal $y[n]$. How can I do this?
I think the solution in the first part of you post is wrong (your solution is shifted by a sample). If I understand correctly, it looks like you are trying to solve the following system:
$$ \begin{array}{l} 3 a + b = -1 \\ 2 a + b = 2 \end{array} $$
which gives $a = -3$ and $b = 8$. With these numbers you'll find that $x[-3n + 8] = y[n]$.
To find $x[4n - 9]$, just substitute $4n-9$ wherever $n$ appears in the definition of $x$:
$$ x[4n-9] = 4\delta [4n +6] +2\delta[4n-4] - 8 \delta[4n-12] $$