Solving $b(5px - 3x) = a(qx - 4)$ for $x$, and stating any restrictions on the variables I am a high school student in Algebra II and while I normally have no trouble with problems dealing with algebraic equations, I simply cannot muster the answer to this question.

Solve for $x$:
$$b(5px - 3x) = a(qx - 4)$$
State any restrictions on the variables.

I'll show you the process of deriving the solution that I have been taking, and if you could identify where I have been erroneous that would be greatly appreciated.
\begin{align*}
b(5px - 3x) &= a(qx - 4)\\
5bpx - 3bx &= aqx - 4a \\
5bpx &= aqx - 4a + 3bc \\
5bpx - aqx &= -4a + 3bc\\
x(5bp - aq) &= -4a + 3bc\\
x &=  \frac{-4a + 3bc}{5bp - aq}&&& (5bp ≠ aq)
\end{align*}

In the between the second and third lines, it looks like you just made a typo. You added $3bx$ to both sides, but then you wrote $3bc$ on the other side. The equation in the third line should have been 
$$ 5bpx = aqx - 4a + 3bx$$
