Balancing a redox reaction with only one product
You should use the the same product species in both half -reactions. In the example you gave, this is $\ce{N2}$.
$$\ce{NH4NO3-&gt; N2}$$
The half-reactions are constructed from the ammonium and nitrate ions separately, since these feature nitrogen in different oxidation states:
\begin{align}
\ce{NH4+ &amp;-&gt; N2} \tag{1} \\
\ce{NO3- &amp;-&gt; N2} \tag{2}
\end{align}
Balancing the reduction half-equation $(1)$:
\begin{align}
\ce{2NO3- &amp;-&gt; N2} &amp; &amp; \text{(balance N)} \\
\ce{2NO3- &amp;-&gt; N2 \color{red}{+ 6H2O}} &amp; &amp; \text{(add } \ce{H2O} \text{ to balance O)} \\
\ce{2NO3- \color{red}{+ 12H+} &amp;-&gt; N2 + 6H2O} &amp; &amp; \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NO3- + 12H+ \color{red}{+ 10e-} &amp;-&gt; N2 + 6H2O} &amp; &amp; \text{(add } \ce{e-} \text{ to balance charge / oxidation state)}
\end{align}
And balancing the oxidation half-reaction:
\begin{align}
\ce{2NH4+ &amp;-&gt; N2} &amp; &amp; \text{(balance N)} \\
\ce{2NH4+ &amp;-&gt; N2 \color{red}{+ 8H+}} &amp; &amp; \text{(add } \ce{H+} \text{ to balance H)} \\
\ce{2NH4+ &amp;-&gt; N2 + 8H+ \color{red}{+ 8e-}} &amp; &amp; \text{(add } \ce{e-} \text{ to balance charge / oxidation state)} \\
\end{align}
In order to combine the two balanced half-equations, we need to multiply the first by $4$ and the second by $5$:
\begin{align}
\ce{8NO3- + 48H+ + 40e- &amp;-&gt; 4N2 + 24H2O} \\
\ce{10NH4+ &amp;-&gt; 5N2 + 40H+ + e-} \\ \hline
\ce{8NO3- + 10NH4+ + 48H+ + 40e- + 10H2O &amp;-&gt; 9N2 + 24H2O + 40H+ + 40e-}
\end{align}
which finally reduces to
$$\ce{8NO3- + 10NH4+ + 8H+ -&gt; 9N2 + 14H2O}.$$