How do you work this thermodynamics question out?

Given :
3C(s) + 4H₂(g) + (1/2)O₂(g) → CH₃CH(OH)CH₃(l) ...... ΔH = -318 kJ
C(s) + O₂(g) → CO₂(g) ...... ΔH = -394 kJ
H₂(g) + (1/2)O₂ → H₂O(l) ...... ΔH = -286 kJ

Rewrite the three thermochemical equations as :
CH₃CH(OH)CH₃(l) →3C(s) + 4H₂(g) + (1/2)O₂(g) ...... ΔH = -(-318) kJ
3C(s) + 3O₂(g) → 3CO₂(g) ...... ΔH = 3(-394) kJ
4H₂(g) + 2O₂ → 4H₂O(l) ...... ΔH = 4(-286) kJ

Add the above three rewritten thermochemical equations, and cancel 3C(s), 4H₂(g) and (1/2)O₂(g) on the both sides.
CH₃CH(OH)CH₃(l) + (9/5)O₂(g) → 3CO₂(g) + 4H₂O(l) ...... ΔH = -2008 kJ

Hence, ΔHc[CH₃CH(OH)CH₃(l)] = -2008 kJ mol⁻¹