This time we choose to eliminate the \(x\)-terms. We must arrange things so that the coefficients of the \(x\)-terms are opposites, so we look for the smallest integer that both 2 and 5 divide into evenly. (This number is called the lowest common multiple, or LCM, of 2 and 5.) The LCM of 2 and 5 is 10. We want one of the coefficients of \(x\) to be \(10\text{,}\) and the other to be \(-10\text{.}\)

To achieve this, we multiply the first equation by \(2\) and the second equation by \(-5\text{.}\)

\begin{align*}
\alert{2}(5x-2y\amp=22)\amp\amp \rightarrow\amp 10x - 4y \amp=44 \\
\alert{-5}(2x-5y\amp=13)\amp\amp \rightarrow\amp -10x +25y \amp=-65
\end{align*}

Adding these new equations eliminates the \(x\)-term and yields an equation in \(y\text{.}\)

\begin{align*}
10x - 4y \amp = \hphantom{-}44\\
\underline{-10x + 25y} \amp =\underline{ -65 \vphantom{y}}\\
21y \amp = -21
\end{align*}

We solve for \(y\) to find \(y = -1\text{.}\) Finally, we substitute \(y = \alert{-1}\) into the first equation and solve for \(x\text{.}\)

\begin{align*}
5x-2(\alert{-1})\amp = 22\\
5x\amp = 20\\
x\amp = 4
\end{align*}

The solution of the system is the point \((4, -1)\text{.}\)