[tex]\boxed{4.}~x+3=2 \Rightarrow x=2-3 \Leftrightarrow x=-1. \\ \\ |3x+y| \leq 1 \Leftrightarrow |-3+y| \leq 1 \Leftrightarrow -1 \leq -3+y \leq 1 \Leftrightarrow 2 \leq y \leq 4 \Rightarrow \\ \\ \Rightarrow y \in \{2,3,4\}. \\ \\ Avem,~deci:~(x,y) \in \{(-1,2);(-1,3);(-1,4) \}.[/tex]
[tex]\boxed{5.}~a)~x^2+5x+6=x^2+2x+3x+6=x(x+2)+3(x+2)= \\ \\ =(x+2)(x+3). \\ \\ b)~ \frac{x^2+5x+6}{x^2+6x+9}= \frac{(x+2)(x+3)}{(x+3)^2}= \frac{x+2}{x+3}=1- \frac{1}{x+3}\ \textless \ 1~(deoarece~ \frac{1}{x+3}\ \textgreater \ 0) . [/tex]
