How do I balance the combustion reaction of butene?
The balanced equation for the combustion of butene is
$$\ce{C4H8 + 6O2 -&gt; 4CO2 + 4H2O}.$$
Nitrogen is included only to indicate that it is present in the medium with some specific ratio to the oxygen amount &ndash; well technically it is included to make question look complicated.

$\ce{C4H8}$ in the products side is included to indicate that not all of it can be combusted but some escapes. And it really is arbitrary, depends on the the engineering of combustion room, etc. &ndash; which is why you have 4 equations, 5 unknowns: d is free to vary.  But you are given the dry analysis to fix its value.

Looking at the equations given, you have

\begin{align}
  b &amp;= c = 4(1-d)\\
  a &amp;= e = b + c/2 = 6(1-d) = 3b/2
\end{align}

And dry analysis says ratio of $\ce{CO2}$:$\ce{C4H8}$ is $14.95:0.75 = 19.93 (b/d)$ after the combustion. And this is the last equation you need.