# Gaussian elimination vs substitution

I'm just starting to learn linear algebra and Gaussian elimination to solve systems of linear equations. In the school I used to solve systems of equations by substitution, e.g. express $Y$ via $X$ and then substitute this expression instead of $Y$. Then find $X$ and then find $Y$ from the first substitution. I'm curious why do we need Gaussian elimination, reduced row echelon forms in order to solve systems of linear equations if "old" approaches work too?

$$\left(\begin{array}{cc|c} 1 & 1 &2 \\ 0 & 1 & 3 \end{array} \right),\iff \left\{\begin{array}{cccc} x&+&y&=2\\ &&y&=3 \end{array} \right.$$
You can see that substitution yields, $x+3=2,$ so $x=-1.$ Let $R_1$ be row 1 and $R_2$ be row 2, then
$$\left(\begin{array}{cc|c} 1 & 1 &2 \\ 0 & 1 & 3 \end{array} \right)\quad \overset{\implies}{R_1-R_2=R_1} \quad \left(\begin{array}{cc|c} 1 & 0 &-1 \\ 0 & 1 & 3 \end{array} \right)\iff \left\{\begin{array}{c} x&=-1\\ y&=~~~3 \end{array} \right.$$ So you can see that we get the same result. The crucial point here is that elimination is a more general form of substitution, where you don't need to know what the variable is explicitly before substituting it into your equations.