Observation 4.3.1.
We can use substitution in a more general way to solve linear systems. For example, a natural approach to the system
\begin{equation*}
\begin{alignedat}{3}
x \amp {} + {} \amp 2y \amp {}={} \amp 1 \\
2x\amp {}+{} \amp 3y \amp {}={} \amp 3. \\
\end{alignedat}
\end{equation*}
is to use the first equation to express \(x\) in terms of \(y\text{:}\)
\begin{equation*}
x = 1-2y
\end{equation*}
and then substitute this into the second equation and simplify:
\begin{equation*}
\begin{alignedat}{2}
2x + 3y \amp {}= \amp 3 \\
2(1-2y) + 3y \amp {}={} \amp 3 \\
2-4y + 3y \amp {}={} \amp 3 \\
-y \amp {}={} \amp 1 \\
y \amp {}={} \amp -1 \\
\end{alignedat}
\end{equation*}
From here, we can substitute \(y=-1\) into the first equation to arrive at the solution \((x,y)=(3,-1)\text{.}\)
However, the two-step process of solving for \(x\) in terms of \(y\) and substituting into the second equation may be performed more efficiently by adding a multiple of the first equation to the second. In this case, we will multiply the first equation by -2 and add to the second equation
\begin{equation*}
\begin{array}{cr}
\amp -2(\text{equation 1}) \\
+ \amp \text{equation 2} \\
\hline
\end{array}
\end{equation*}
to obtain
\begin{equation*}
\begin{array}{cr}
\amp -2(x+2y=1) \\
+ \amp 2x+3y = 3 \\
\hline
\\
\end{array}
\end{equation*}
which gives us
\begin{equation*}
\begin{array}{crcr}
\amp -2x-4y \amp = \amp -2 \\
+ \amp 2x+3y \amp = \amp 3 \\
\hline
\amp -y \amp = \amp 1. \\
\end{array}
\end{equation*}
In this way, the system
\begin{equation*}
\begin{alignedat}{3}
x \amp {} + {} \amp 2y \amp {}={} \amp 1 \\
2x\amp {}+{} \amp 3y \amp {}={} \amp 3 \\
\end{alignedat}
\end{equation*}
is transformed into the new triangular system
\begin{equation*}
\begin{alignedat}{3}
x \amp {} + {} \amp 2y \amp {}={} \amp 1 \\
\amp \amp -y \amp {}={} \amp 1. \\
\end{alignedat}
\end{equation*}
Of course, the choice to multiply the first equation by -2 was made so that the terms involving \(x\) in the two equations will cancel leading to a triangular system that can be solved using back substitution.
