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I know that to solve a linear equation involving $n$ variables for example, we need $n-1$ other independent equations to form a system and then solve that system. Is there a formal proof for this?

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  • $\begingroup$ It depends on the type of equation. Are they all linear? Occurring over a field? $\endgroup$
    – Randall
    Feb 21, 2019 at 18:32
  • $\begingroup$ I'm talking about linear equations $\endgroup$
    – GDGDJKJ
    Feb 21, 2019 at 18:33
  • $\begingroup$ The solution can only be unique , if we have at least as many equations than variables. This can be shown by looking at the poosible ranks of the coefficient matrix. $\endgroup$
    – Peter
    Feb 21, 2019 at 18:44

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Proof: if we didn't have n equations for n unknowns, then the matrix representing the system would either be rectangular (meaning no unique solution) or have zero rows (also no unique solutions). Therefore there must be n equations for n unknowns in order for a unique solution to exist.

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If it's a system of linear equations, then you can use Gauss-Jordan elimination to get the solutions. You can also take the determinant of the coefficient matrix first to check whether or not you'll have a single solution, infinitely many solutions, or zero solutions.

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