Are there ways to solve equations with multiple variables? I am not at a high level in math, so I have a simple question a simple Google search cannot answer, and the other Stack Exchange questions does not either. I thought about this question after reading a creative math book. Here is the question I was doing, which I used the solutions manual in shame(Not exact wording, but same idea):
The question in the blockquotes below is not the question I am asking for answers to. Some misunderstood what I am asking. What I am asking is in the last single-sentence paragraph.

Suppose $a_2,a_3,a_4,a_5,a_6,a_7$ are integers, where $0\le a_i< i$.
$\frac 57 = \frac {a_2}{2!}+\frac {a_3}{3!}+\frac {a_4}{4!}+\frac {a_5}{5!}+\frac {a_6}{6!}+\frac {a_7}{7!}$
Find $a_2+a_3+a_4+a_5+a_6+a_7$.

The solution to this particular question requires that $a_7$ and the other variables in later steps of the algebra process to be remainders when both sides are to be divided by an integer. I am now wondering what if an equation ever comes up where the method to solve for the question above cannot work due to the variables not being able to return remainders. Thus, my question is whether it is possible to solve algebraic equations with more than two variables and most variables having constant coefficients, is not a system of equations, and the variables are assumed to be integers, and the solution is unique.
Does such a way to solve such equations in general exist? If so, please explain. What is this part of math called?
Thank you.

Multiplying by the common denominator $7!$, we get
$$ 2520 a_2 + 840 a_3 + 210 a_4 + 42 a_5 + 7 a_6 + a_7 = 3600$$
Take this mod $7$:
$$ a_7 \equiv 2 \mod 7$$
Since $0 \le a_7 < 7$, this means $a_7 = 2$, and then (substituting this and dividing by $7$):
$$ 360 a_2 + 120 a_3 + 30 a_4 + 6 a_5 + a_6 = 514$$
Taking this mod $6$:
$$ a_6 \equiv  4 \mod 6$$
So $a_6 = 4$.
Continuing the process, we get 
$$ a_5 = 0,\ a_4 = 1,\ a_3 =  1,\ a_2 = 1 $$
