\(\dfrac{x+1}{35}+\dfrac{x+3}{33}=\dfrac{x+5}{31}+\dfrac{x+7}{29}\\ \dfrac{x+36}{35}+\dfrac{x+36}{33}=\dfrac{x+36}{31}+\dfrac{x+36}{29}\\ \dfrac{x+36}{35}+\dfrac{x+36}{33}-\dfrac{x+36}{31}-\dfrac{x+36}{29}=0\\ \left(x+36\right)\left(\dfrac{1}{35}+\dfrac{1}{33}-\dfrac{1}{31}-\dfrac{1}{29}\right)=0\\ x+36=0\\ x=-36\)