Given: $$61 = (35 + 90x) / a$$ $$610 = (1000 + 20000x) / b$$
Is there a way to determine a value of a, b, and/or x?
It seems like, if I could somehow calculate $a*x$ and $b*x$, perhaps by moving them to a side by themselves, I can narrow down x to be one of their common divisors.
$$a*x = 300$$ $$b*x = 15$$
Common Divisors Of 300 and 15 = 15, 5, 1, 3 <-- value of x
I can't seem to re-arrange the equation to produce this effect though.
What is the best way to solve for a, b, and/or x? Examples please, my math is a bit rusty.
As you write about common divisors, I assume that $a$, $b$, and $x$ are all meant to be integers. After some minor simplifications, you have:
$$61a = 35+90x\\ 61b = 100+2000x$$
Working modulo $61$ both $a$ and $b$ will drop out of the equations:
$$0 = 35+90x \mod 61\\ 0 = 100+2000x \mod 61$$
From the second of these we get: $$0 = 1+20x \mod 61\\ 0 = 3+60x \mod 61\\ -60x = 3 \mod 61\\ x = 3 \mod 61$$
Luckily $x=3$ is also a solution to the first modular equation ($35+90*3=305=5*61$ so it is $0$ modulo $61$), so the two equations are compatible.
So now we know that $x=61k+3$ for some integer $k$. If you substitute that into the original equations, you get $a = 90k+5$ and $b = 2000k+100$. So for any integer value of $k$, you have the solution: $$(a,b,x) = (90k+5, 2000k+100, 61k+3)$$