Using a MS Excel spreadsheet, was conjecturing the following scenario, of a congruence such that:
$ 7+3*x\equiv2*x+5\pmod{42}$
Where a solution is:
$\quad x = 6$
In 'rebalancing' the congruent equation to:
$ 12+5*x\equiv0\pmod{42}$
We also obtain a solution of:
$\quad x = 6$
However in algebraically 'rebalancing' the first congruent equation to:
$ 2+x\equiv0\pmod{42}$
The solutions obtained were:
$\quad x = {0,1,4,5,12, 19...}$
Is the first method of 'rebalancing' mathematically sound?
Or was I engaging in 'Voodoo' mathematics?
As Bill Dubuque commented, your first "rebalancing" was not valid.
$7+3x\equiv2x+5\mod42\implies(7+3x)-(2x+5)=2+x\equiv0\mod42$.
Furthermore, the solution to $2+x\equiv0 \mod 42$ is $x\equiv-2\equiv40\mod 42$.
I don't know how you got $x=0, 1, 4, 5, 12, 19... $, but that would imply $x+2\equiv 2, 3, 6, 7, 14, 21... $,
respectively, not $x+2\equiv0\mod 42$.
Solving equations in integers modulo $42$ is similar to solving equations in $\Bbb R$.