A(a; b+1) ∈ Gf ∪ Gg;
f(x) = 2x + 5;
g(x) = x + 1;
f(a) = 2a + 5;
g(a) = a + 1;
(b + 1) = f(a) = g(a);
⇒ 2a + 5 = a + 1;
2a = a - 4;
a = -4;
b + 1 = 2a + 5;
b + 1 = 2*(-4) + 5;
b + 1 = -8 + 5;
b + 1 = -3;
b = -4;
b + 1= a + 1;
b + 1 = -4 + 1;
b + 1 = -3;
b = -4;
⇒ A(-4; -3);
a = -4;
b = -4;