Question

Помогите решить уравнения (1+c)*7=21 (k-3)*6=48
СРОЧНО!!!

Answer 1

Ответ:

1.с=2

2.с=33/3(или 11)

Answer 2

$\displaystyle \tt (1+c)\cdot7=21\\\displaystyle \tt 7(1+c)=21\\\displaystyle \tt 7+7c=21\\\displaystyle \tt 7c=21-7\\\displaystyle \tt 7c=14\\\displaystyle \tt c=14\div7\\\displaystyle \tt \bold{c=2}$

$\displaystyle \tt (k-3)\cdot6=48\\\displaystyle \tt 6(k-3)=48\\\displaystyle \tt 6k-18=48\\\displaystyle \tt 6k=48+18\\\displaystyle \tt 6k=66\\\displaystyle \tt k=66\div6\\\displaystyle \tt \bold{k=11}$