We first decide which variable to eliminate, \(x\) or \(y\text{.}\) We can choose whichever looks easiest. In this problem, we choose to eliminate \(x\text{.}\)
We next look for the smallest number that both coefficients, \(2\) and \(3\text{,}\) divide into evenly. This number is \(6\text{.}\) We want the coefficients of \(x\) to become \(6\) and \(-6\text{,}\) so we will multiply Equation (1) by \(3\) and Equation (2) by \(-2\) to obtain
\begin{alignat*}{3}
6x\amp {}+{}\amp 9y\amp ={}\amp 24 \hphantom{blankblank}\amp(1\text{a})\\
-6x\amp {}+{}\amp 8y\amp ={}\amp 10 \hphantom{blankblank}\amp(2\text{a})
\end{alignat*}
Now we add the corresponding terms of (1a) and (2a). The \(x\)-terms are eliminated, yielding an equation in one variable.
\begin{alignat*}{3}
6x\amp {}+{}\amp 9y\amp ={}\amp 24 \hphantom{blankblank}\amp(1\text{a})\\
-6x\amp {}+{}\amp 8y\amp ={}\amp 10 \hphantom{blankblank}\amp(2\text{a})\\
{}\amp {}{}\amp 17y\amp ={}\amp 34 \hphantom{blankblank}\amp(3)
\end{alignat*}
We solve this equation for \(y\) to find \(y=\alert{2}\text{.}\) We can substitute this value of \(y\) into any of our equations involving both \(x\) and \(y\text{.}\) If we choose Equation (1), then
\begin{equation*}
2x + 3 (\alert{2}) = 8
\end{equation*}
and solving this equation yields \(x=1\text{.}\) The ordered pair \((1, 2)\) is a solution to the system. You should verify that these values satisfy both original equations.
