Finding the sum of coefficients of the reaction
Linear equations aside, note that this is a redox reaction and you can use half-reactions method as well (BTW the reaction looks fine to me):

$$\ce{$x$ \overset{0}{Zn} + y H\overset{+5}{N}O3 -&gt; $a$ \overset{+2}{Zn}(NO3)2 + b H2O + $c$ \overset{-3}{N}H4NO3}$$

$$
\begin{align}
\ce{\overset{+5}{N} + 8 e- &amp;→ \overset{-3}{N}} \tag{red}\\
\ce{\overset{0}{Zn} &amp;→ \overset{+2}{Zn} + 2 e-} &amp;|\cdot 4 \tag{ox}\\
\hline
\ce{4 \overset{0}{Zn} + \overset{+5}{N} &amp;→ 4 \overset{+2}{Zn} + \overset{-3}{N}} \tag{redox}
\end{align}
$$

So that the complete balanced equation is

$$\ce{4 Zn + 10 HNO3 -&gt; 4 Zn(NO3)2 + 3 H2O + NH4NO3}$$

leaving $a = 4$, $b = 3$, $c = 1$ and the sum $a + b + c = 4 + 3 + 1 = 8$.