t: Tretboote, r: Ruderboote, k: Kanus

${\displaystyle {\begin{array}{rcrcrcl}t&+&r&+&k&=&43\ {\text{Boote insgesamt}}\\5\ t&+&3\ r&+&2\ k&=&159\ {\text{Personen höchstens}}\\8\ t&+&7\ r&+&4\ k&=&271\ \mathrm {\euro} \ {\text{höchstens pro Stunde}}\end{array}}}$

Erst Lösbarkeit mit Determinante überprüfen:

${\displaystyle A={\begin{vmatrix}1&1&1\\5&3&2\\8&7&4\\\end{vmatrix}}={\begin{vmatrix}{\bcancel {1}}&{\bcancel {1}}&{\overset {-24}{\xcancel {1}}}\\5&{\xcancel {3}}&{\xcancel {2}}\\{\cancel {8}}&{\cancel {7}}&{\underset {+12}{\xcancel {4}}}\\\end{vmatrix}}{\begin{vmatrix}{\overset {-14}{\cancel {1}}}&{\overset {-20}{\cancel {1}}}\\{\xcancel {5}}&3\\{\underset {+16}{\bcancel {8}}}&{\underset {+35}{\bcancel {7}}}\\\end{vmatrix}}=\dots }$

${\displaystyle \dots =12+16+35-24-14-20=-3\neq 0}$

Das System hat eine Lösung:

${\displaystyle \left({\begin{array}{c c c | c}1&1&1&43\\5&3&2&159\\8&7&4&271\\\end{array}}\right){\begin{array}{c }\cdot (-2)in\ 2.\\\cdot (-2)in\ 3.\\\\\end{array}}\quad \left({\begin{array}{c c c | c}1&1&1&43\\3&1&0&73\\-2&1&0&-47\\\end{array}}\right){\begin{array}{c }\\\cdot (-1)in\ 3.\\\\\end{array}}\quad }$

${\displaystyle \left({\begin{array}{c c c | c}1&1&1&43\\3&1&0&73\\-5&0&0&-120\\\end{array}}\right){\begin{array}{c }\\\\:(-5)dann\cdot (-3)in\ 2.\\\end{array}}}$

${\displaystyle \left({\begin{array}{c c c | c}1&1&1&43\\0&1&0&1\\1&0&0&24\\\end{array}}\right){\begin{array}{c }\\\cdot (-1)in\ 1.\\\cdot (-1)in\ 1.\\\end{array}}\quad \left({\begin{array}{c c c | c}0&0&1&18\\0&1&0&1\\1&0&0&24\\\end{array}}\right)}$

Also 24 Tretboote, 1 Ruderboot und 18 Kanus.