Determine $a$, $b$ and $n$ in the following expression (binomial theorem backwards) You are given the following expression:

$$(ax + by)^{n} = -15120x^{4}y^{3}$$

Determine the constants $a$, $b$ and $n$.

My attempt to solve this problem is by trying to use the binomial theorem backwards.

The binomial theorem: $(ax + by)^{n} = \displaystyle\sum_{k = 0}^{n} {\binom{n}{k}(ax)^{n-k}(by)^{k}}$

And so if you compare the expressions, you get: 

$$\binom{n}{k}(a)^{n-k}(b)^{k} = -15120$$

$n - k = 4$ and $k = 3$ so that $n = 7$

Here is where I get stuck, because now I have:

$$\binom{7}{3}(a)^{4}(b)^{3} = -15120$$

Two unknowns... How to solve it? Am I even doing it correctly?
I think, that the exercise is


  You are given the following expression:
  
  $(ax + by)^{n} = \color{blue}{\ldots} -15120x^{4}y^{3}+
 \color{blue}{\ldots} $
  
  Determine the constants $a$, $b$ and $n$.


In this case you know, that the 5th summand is ${7 \choose 4} \cdot (ax)^4\cdot (by)^3=-15120x^{4}y^{3}$

$15120=2^4\cdot3^3\cdot5\cdot7$.

${7 \choose 4}=5\cdot 7$

Thus $b=-3$ and $a=2$
You get $b= (\frac{-15120}{35 \, a^{4}})^{\frac{1}{3}}$ , $a$ can be any value as we only have one term.