tolong dibantu dengan cara penyelesaiannya

[tex]f(x) \left \{ {{-x-2; x\ \textless \ 0} \atop {x-2;x \geq 0}} \right. [/tex]
[tex]g(x) \left \{ {{x+2;x\ \textless \ 0} \atop {-x;x \geq 0}} \right. [/tex]
[tex]\therefore -x-2 - x-2 = -1 \vee x-2 +x =-1[/tex]
[tex]\therefore x_1 = -\frac{3}{2} \vee x_2 =\frac{1}{2}[/tex]
[tex] x_1+x_2 = -1 [/tex]