Finding the values of $A,B,C,D,E,F,G,H,J$ Given that the letters $A,B,C,D,E,F,G,H,J$ represents a distinct number from $1$ to $9$ each and 
$$\frac{A}{J}\left((B+C)^{D-E} - F^{GH}\right) = 10$$
$$C = B + 1$$
$$H = G + 3$$
find (edit: without a calculator) $A,B,C,D,E,F,G,H,J$
I could only deduce that $D\ge E$, from the first one. Eliminating C and H doesn't seem to help much either.

We have that $J|A$, and that $\frac{A}{J}|10$. So let's first consider the possible divisors of $10$, which are $1, 2, 5$. Clearly, $\frac{A}{J} \neq 5$ is the most likely option, based on the possible values. So $\frac{A}{J} = 2$. How many ways can we get this? Consider pairs $(A, J)$. We have $(2, 1)$, $(4, 2)$, $(8, 4)$, $(6, 3)$. 
We now have that since $\frac{A}{J} = 2$ that $(2B + 1)^{D - E} - F^{G^{2} + 3G} = 5$. I went ahead and substituted based on the constraints given. It will be most helpful to look at how the various digits behave under modular exponentiation, using modulo 10. So for example, when we exponentiate $3$, we get the one's place as $3^{1} \to 3$, $3^{2} \to 9$, $3^{3} \to 7$, $3^{4} \to 1$, $3^{5} \to 3$. The minimum value such that $a^{x} \equiv 1$ $(mod$ $10)$ is called the order of $a$ modulo 10. Once you pick the elements, it comes down to making sure the exponents are in line. Noting the order of an element will help you here.
So how can you make $5$ on the digits places? You have $6 - 1$, $8 - 3$, and $9 - 4$.
I think this should be a sufficient hint to get you going in the right direction. 
