Can someone walk me through this wolframalpha solution? Given
$$x = (625/50)a-(100/50)$$
$$x = (214/60)b-(110/60)$$
The following integer solution was provided by a wolframalpha widget:
$$a = 214n+10$$
$$b = 750n+35$$
$$x=2675n+123$$
I am very interested in how the 10, 35, and 123 were determined here because they are the actual values of a, b and x and I didn't think they couldn't be solved.
This is the only algebra solver I found that produces these values and unfortunately they don't show a step by step here.
Can anyone walk me through the steps that produce this solution? Here is a link to the wolframalpha widget.
 A: Wolfram also gives a solution in the reals which is equivalent to the following:
$$b = \frac{5(75 a - 1)}{107}\\x = \frac{25 a-4}{2}$$
This is what you get if you consider $a$ a constant and solve for the two remaining variables $x$ and $b$ in terms of $a$.
We want everything to be integers. This is easy for second equation - for $x$ to be an integer, we need $a$ to be an even number.
The first equation is trickier - for $b$ to be an integer, we need $75 a - 1$ to be a multiple of $107$. By trying various small numbers for $a$, you will find soon enough that $a=10$ works: $75*10-1 = 749 = 7*107$. Once you have that solution, then you also know that $a=10+107$ will work to make $b$ an integer, as will $a=10+2*107$, or in fact $a=10+107k$ for any integer $k$.
Combining this with the fact that $a$ must be even, you get that $k$ is even, so $a=214n+10$. The expressions for $b$ and $x$ then follow by substituting this into the equations.
I skipped a lot of details in the above to give you the overall flow of the reasoning. Those details have to do with Modular Arithmetic. Here is just a short overview of those details, with links for you to find out more.
1)  When is $75 a - 1$ a multiple of $107$?
This is just the modular equation $75a-1\equiv0 \mod 107$, or $a\equiv 75^{-1} \mod 107$. To solve it, you need to find the inverse of $75$ modulo $107$, which exists because $75$ and $107$ are coprime. This is done using the Extended Euclidean Algorithm, which gives the result that $10*75-7*107=1$, so $10$ and $75$ are inverses modulo $107$.
2) How do you combine the restrictions on $a$ imposed by the two equations?
In this case the second equation imposed a simple restriction, that $a$ is even. We actually have the two modular equations:
$$a \equiv 10 \mod 107\\a \equiv 0 \mod 2$$
With different numbers it would have been a bit trickier to combine them into a single restriction, and we would have had to use the Chinese Remainder Theorem.
A: Subtracting,
$$\frac{25}{2}a-2-\frac{107}{30}b+\frac{11}{6}=0$$
Get rid of the denominators
$$15\cdot25a-60-214b+55=0$$
$$375a-214b-5=0$$
It's now obvious that it's a usual linear diophantine equation. <-- As you can see here, it can be solved with the extended Euclidean algorithm.
