Question

Решите 3 и 4 пж. Дам 15 балов

Answer 1

$\displaystyle 1)39k+(20k-k)=6496\\39k+(19k)=6496\\39k+19k=6496\\58k=6496\\k=112\\\\ 2)(c:22)+11+933=1800\\\\ (\frac{1}{22}c)+11+933=1800\\\\ \frac{1}{22}c+11+933=1800\\\\ \frac{1}{22}+944=1800\\\\22(\frac{1}{22}c+944)=22*1800\\\\ 22*\frac{1}{22}c+22*944=39600\\\\ c+22*944=39600 \\ c+20768=39600\\c=39600-20768\\c=18832$