Need help I really don't understand this Chemistry!!!!!!!!!

Calculate the standard reaction enthalpy for the reaction.
CH4(g) + H2O(g) = CO(g) + 3H2(g)

given

2H2(g) + CO(g) = CH3OH(l) delta H degrees = −128.3 kJ · mol−1
2CH4(g) + O2(g) = 2CH3OH(l) delta H degrees = −328.1 kJ · mol−1
2H2(g) + O2(g) = 2H2O(g) delta H degrees = −483.6 kJ · mol−1

1. +412.1 kJ · mol−1
2. +155.5 kJ · mol−1
3. +216 kJ · mol−1
4. +206.1 kJ · mol−1
5. +42.0 kJ · mol−1

Please help...I'm totally lost!!!!

Explain!!!!!!! Thank you!!

Your target reaction is CH4(g) + H2O(g) = CO(g) + 3H2(g)

Do whatever you must with your GIVENS (reverse equation, divide, multiply, W/E) to get to your target reaction. **What ever you do to your equation, do with same with your enthalpy. Thus..

Let's label the given reactions A, B, and C respectively.

A) 2H2(g) + CO(g) = CH3OH(l) delta H degrees = −128.3 kJ · mol−1
B) 2CH4(g) + O2(g) = 2CH3OH(l) delta H degrees = −328.1 kJ · mol−1
C) 2H2(g) + O2(g) = 2H2O(g) delta H degrees = −483.6 kJ · mol−1

Now to get to the target, lets start by obtaining 1CH4....so we'll divide reaction "B" by two to lower the coefficient of all elements/compounds in equation by the division of "2" thus..

B/2 ) CH4(g) + 1/2 O2(g) = CH3OH(l) delta H degrees = −328.1 / 2 ---> -164.05

Now let's get one H2O (H20 is in equation "C"). We will reverse the equation C and divide by two b/c the equation in the target reaction has H2O in reactants side and mole mole of it.

-C/2 ) H2O = H2 + 1/2 O2 delta H degrees = - (−483.6/ 2) ---> 241.8

Now let's get Now we have to get CO. Equation A has CO on reactants but our target has it on the products side, (and moles/coefficient is same number) so let's flip it...

-A) CH3OH = CO + 2H2 delta H degrees = - (−128.3) ---> 128.3

So let's show all the altered givens in clear view...

B/2 ) CH4(g) + 1/2 O2(g) = CH3OH(l) delta H degrees = −328.1 / 2 ---> -164.05
-C/2 ) H2O = H2 + 1/2 O2 delta H degrees = - (−483.6/ 2) ---> 241.8
-A) CH3OH = CO + 2H2 delta H degrees = - (−128.3) ---> 128.3

Now we can cancel out the 1/2 O2 (from the reactant side of B/2 and from the Product side of -C/2). Also, the CH3OH (from B/2 and -A). Now, since we have one H2 on products side in -C/2 and 2 H2 on products side ALSO in -A, we can add them to get 3H2 (which is what the target has). Now we have everything trimmed out to the target equation. Now that everything is correct, just add the altered enthalpies.

-164.05 + 241.8 + 128.3 = 206.05 or about 206.1kJ mol-1 WHICH IS answer choice "4."