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So I have a set of equations:

$$x_{1} + x_{2} = 1$$ $$x_{2} + x_{4} = 3$$

From linear algebra, we know that (say, we're in $\mathbb{R}^{4}$, i.e. we have 4 variables), the solution space to the set of equations above is a 2 dimensional subspace of $\mathbb{R}^{4}$.

But now, using algebra we could also do the following simplification(...?)

$$ - x_{1} + x_{4} = 2$$

Clearly, the solution space becomes different! (try, x1 = 1, x2 = 3, x3 = 0, x4 = 3)

From a linear algebra perspective, what is really going on here. I am not grasping this intuitively. Thanks a lot. (I think I am missing something about correctly reducing a set of equations...)

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    $\begingroup$ From $\begin{cases} x_1+x_2&=1\\ x_2+x_4&=3\end{cases}$ you correctly deduced that $-x_1+x_4=2$, in symbols $$\begin{cases} x_1+x_2&=1\\ x_2+x_4&=3\end{cases}\implies -x_1+x_4=2.$$ You then provide an example that satisfies the consequent but not the antecedent. So what? The set that satisfies the system isn't equal to $\{(x_1, x_2, x_3, x_4)\colon -x_1+x_4=2\}$, this is true, but the single equation system $-x_1+x_4=2$ isn't equivalent to the starting system either, so they will have different solutions. $\endgroup$ – Git Gud Jun 29 '15 at 22:06
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Your new equation is not a simplification of the system, it's just a linear combination of the original equations.

Substitution is a way of deriving a new equation that follows logically from a system. An equation created by substitution won't in general define the system.

If you take your new equation plus any one of the original equations, you'll find it defines the same region as the original system.

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  • $\begingroup$ Great. You've cleared it up for me. Thanks a lot! $\endgroup$ – Andy Yao Jun 30 '15 at 1:33
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The operation you are doing is called elimination. The new equation created from elimination is linearly dependent on the rows added or subtracted to create it, so it will need to replace one of the rows if you want your system to maintain linear independence.

So your new system would either be:$$ x_1+x_2=1\\ -x_1+x_4=2 $$Or alternatively:$$ x_2+x_4=3\\ -x_1+x_4=2 $$

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