# Grasping “Substitution” in terms of linear algebra

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...)

• 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. – Git Gud Jun 29 '15 at 22:06

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