Distinct equations and rules to solve for unknowns?

[yo4561]

I learned that if you have three unknown variables and three distinct equations that you can solve for all three unknown variables. Does this rule apply to e.g. five unknown variables and five distinct equations, six unknown variables and six distinct equations etc.? Also to clarify, by distinct does this mean that the equations are not just reworded (e.g. x+y=10 is the same as x=10-y)? How do you know if an equation would even be helpful to allow for you to use it? This rule seems to have some exceptions, no?

[esledge]

I learned that if you have three unknown variables and three distinct equations that you can solve for all three unknown variables. Does this rule apply to e.g. five unknown variables and five distinct equations, six unknown variables and six distinct equations etc.?

Yes, though I would add that they need to be distinct linear equations. Terms with exponents or with products of variables can have more solutions: For example, even though this is one equation with one unknown, x^2 - 5x + 6=0 has two solutions.

Also to clarify, by distinct does this mean that the equations are not just reworded (e.g. x+y=10 is the same as x=10-y)?

Yes, that is what "distinct" means in this context.

How do you know if an equation would even be helpful to allow for you to use it? This rule seems to have some exceptions, no?

That's a loaded question! Sometimes similarity between equations is the clue that they might be useful (see the example below). The two main exceptions are:
(1) Equations that at first appear to be distinct, but upon simplification you see that they actually aren't. (You have less info than it seems at first.)
(2) A question asks for a combo of variables, or combos of variables can be eliminated to leave the one variable you need. (You have more info than it seems at first.)

The 2nd exception is all about how the GMAT won't make you solve for all of the variables in the system of equations (as you did in school). So for example:

What is the value of x?
(1) x + 3z = 2y - 4
(2) 6z - 2y = 8

By the rule, this has 3 variables and only 2 equations, so you can't solve for all three variables. But you can solve for x if you combine the statements, so the answer is (C).