Calculation of the enthalpy change of reactions First, I’m a high school student and I wonder why every textbook just tells you how to set up those equations and cancel them to get the entropy change but never tells you why you can cancel the same species on both sides. 

Second, for example, 

$\ce{2B + 3H2O -&gt; B2H6 + 3/2O2} \tag{  $\Delta H= 762~\mathrm{kJ}$}$ 

$\ce{H2 + 1/2O2 -&gt; H2O} \tag{  $\Delta H=-286~\mathrm{kJ}$}$ 

Here the textbook tells us we should add up two reactions in order to get what we want (I didn’t post the whole question, just go with this)

Now, why do we have to add the two whole reactions up instead of just change the $\ce{H2O}$ on the left side into $\ce{H2 + 1/2O2}$?  What I understand about this $\ce{H2 + 1/2 O2 -&gt;H2O}$ is if we reverse it, we can say if $\ce{H2O}$ takes in $286~\mathrm{kJ}$ of energy, it will turn into $\ce{H2}$ and $\ce{1/2 O2}$. So why we can’t just add $286~\mathrm{kJ}$ to the equation? Why we still need to add the whole thing $\ce{H2 + 1/2 O2 -&gt; H2O}$ to the other equation?
You always need to add the equations since the break of bonds or the formation of bonds leads to an exchange in energy.
Here is a simple case 

$$\begin{alignat}{2}
\ce{2 B + 3 H2O &amp;-&gt; B2H6 + 3/2 O2}\quad&amp;&amp;(\Delta H = 762\ \mathrm{kJ})\\
\ce{H2 + 1/2 O2 &amp;-&gt; H2O}\quad&amp;&amp;(\Delta H = −286\ \mathrm{kJ}) \times 3\\
\hline
\ce{3 H2 + $\boldsymbol{\ce{3/2 O2}}$ &amp;-&gt; $\boldsymbol{\ce{3H2O}}$}\quad&amp;&amp;(\Delta H = −858\ \mathrm{kJ})\\
\ce{2 B + $\boldsymbol{\ce{3 H2O}}$ &amp;-&gt; B2H6 + $\boldsymbol{\ce{3/2 O2}}$}\quad&amp;&amp;(\Delta H = 762\ \mathrm{kJ})
\end{alignat}$$

Notice the ones bold get canceled since the product is the reactant in the other equation.

$\Delta H_\mathrm{r}$ = sum of the $\Delta H$s so it’s equal to $-96\ \mathrm{kJ}$.

And in the second equation you were talking about one $\ce{H2O}$ molecule however in the first you are talking about three.