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

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

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