Die Lösung in der Koordinatenform lautet

$\begin{array}{c}\left(\begin{array}{c}5\\ 11\end{array}\right)=\left(\begin{array}{c}x\\ y\end{array}\right)+\left(\begin{array}{c}3\\ 7\end{array}\right)\\ \\ 5=x+3\\ x=2\\ \\ 11=y+7\\ y=4\end{array}\backslash def\backslash arraystretch\{1.25\}\; \backslash begin\{array\}\{l\}\backslash begin\{pmatrix\}5\backslash \backslash 11\backslash end\{pmatrix\}=\backslash begin\{pmatrix\}x\backslash \backslash y\backslash end\{pmatrix\}+\backslash begin\{pmatrix\}3\backslash \backslash 7\backslash end\{pmatrix\}\backslash \backslash \backslash \backslash 5=x+3\backslash \backslash x=2\backslash \backslash \backslash \backslash 11=y+7\backslash \backslash y=4\backslash end\{array\}$

Der Punkt $PP$ hat die Koordinaten $(2\mathrm{\mid }4)(2|4)$.