Question

  2a+6       ∈ Z    a∈Z  a= ?
a²+4a=3

Answer 1

$\frac{2(a+3)}{(a+1)(a+3)} = \frac{2}{a+1}$
a+1 I2
a+1∈D2 (multimea divizorilor lui 2)
D2={-1,+1,-2,+,-2}
a+1=-1⇒a=-2
a+1=+1⇒a=0
a+1=-2⇒a=-3
a+1=2⇒a=1

Answer 2

nu uita sa pui conditia:
$a+1 \neq 0 \\ a+3 \neq 0$
deci :
$a \neq -1 \\ a \neq -3$