The substitution method in Exercise 51 leads to a quadratic equation. Here is
an alternative approach to solving that system; this approach leads to linear
equations. Multiply equation (2) by 2 and add the resulting equation to
equation (1). Now take square roots to conclude that \(x+y=\pm 3 .\) Next,
multiply equation (2) by \(2,\) and subtract the resulting equation from
equation (1). Take square roots to conclude that \(x-y=\pm 1 .\) You now have
the following four linear systems, each of which can be solved (with almost no
work) by the addition-subtraction method.
$$\begin{array}{ll}\left\\{\begin{array}{l}x+y=3 \\\x-y=1\end{array}\right. &
\left\\{\begin{array}{l}x+y=3 \\\x-y=-1\end{array}\right.
\\\\\left\\{\begin{array}{l}x+y=-3
\\\x-y=1\end{array}\right.\end{array}\left\\{\begin{array}{l}x+y=-3
\\\x-y=-1\end{array}\right.$$ Solve these systems and compare your results
with those obtained in Exercise 51