Calculation of the enthalpy change of reactions
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.